Page 22.4 Ex. 22.1
Q1.
Answer :
In this differential equation, the order of the highest order derivative is 3 and its power is 1. So, it is a differential equation of order 3 and degree 1.
It is a non-linear differential equation because the differential coefficient dxdt has exponent 2, which is greater than 1.
Q2.
Answer :
In this differential equation, the order of the highest order derivative is 2 and its power is 1. So, it is a differential equation of order 2 and degree 1.
It is a linear differential equation.
Q3.
Answer :
dydx2+1dydx=2
⇒dydx3+1dydx=2
⇒dydx3+1=2dydx
⇒dydx3-2dydx+1=0
In this equation, the order of the highest order derivative is 1 and its highest power is 3. So, it is a differential equation of order 1 and degree 3.
It is a non-linear differential equation because the differential coefficient dydx has exponent 3, which is greater than 1.
Q4.
Answer :
1+dydx2=cd2ydx213Squaring both sides, we get⇒1+dydx2=cd2ydx223Taking cubes of both sides, we get⇒cd2ydx22=1+dydx23⇒c2d2ydx22=1+3dydx2+3dydx4+dydx6
In this differential equation, the order of the highest order derivative is 2 and its power is 2. So, it is a differential equation of order 2 and degree 2.
It is a non-linear differential equation, as its degree is more than 1.
Page 22.5 Ex. 22.1
Q5.
Answer :
d2ydx2+dydx2+xy=0
In this differential equation, the order of the highest order derivative is 2 and its power is 1. So, it is a differential equation of order 2 and degree 1.
It is a non-linear differential equation, as the differential coefficient dydx has exponent 2, which is greater than 1.
Q6.
Answer :
d2ydx23=dydx⇒d2ydx213=dydx12Taking cubes of both the sides, we get⇒d2ydx2=dydx32Squaring both the sides, we get⇒d2ydx22=dydx3⇒d2ydx22-dydx3=0
In this differential equation, the order of the highest order derivative is 2 and its power is 2. So, it is a differential equation of order 2 and degree 2.
Thus, it is a non-linear differential equation, as its degree is 2, which is greater than 1.
Q7.
Answer :
d4ydx4=c+dydx232Squaring both sides, we get⇒d4ydx42=c+dydx23⇒d4ydx42=c3+3c2dydx2+3cdydx4+dydx6
In this differential equation, the order of the highest order derivative is 4 and its power is 2. So, it is a differential equation of order 4 and degree 2.
Thus, it is a non-linear differential equation, as its degree is 2, which is greater than 1.
Q8.
Answer :
x+dydx=1+dydx2⇒x+dydx=1+dydx212Squaring both sides, we get⇒x+dydx2=1+dydx2⇒x2+2xdydx+dydx2=1+dydx2⇒2xdydx+x2=1
In this differential equation, the order of the highest order derivative is 1 and the power is 1. So, it is a differential equation of order 1 and degree 1.
Hence, it is a linear differential equation.
Q9.
Answer :
yd2xdy2=y2+1
In this differential equation, the order of the highest order derivative is 2 and its power is 1. So, it is a differential equation of order 2 and degree 1.
It is a linear differential equation.
Q10.
Answer :
s2d2tds2+stdtds=s⇒sd2tds2+tdtds=1
In this differential equation, the order of the highest order derivative is 2 and its power is 1. So, it is a differential equation of order 2 and degree 1.
It is a non-linear differential equation, as it contains the product of the dependent variable t and its differential co-efficient dtds.
Q11.
Answer :
x2d2ydx23+ydydx4+y4=0
In this differential equation, the order of the highest order derivative is 2 and its power is 3. So, it is a differential equation of order 2 and degree 3.
It is a non-linear differential equation, as its degree is more than 1.
Q12.
Answer :
d3ydx3+d2ydx23+dydx+4y=sin x
In this differential equation, the order of the highest order derivative is 3 and its power is 1. So, it is a differential equation of degree 3 and order 1.
It is a non-linear differential equation, as its differential co-efficient
d2ydx2
has exponent 3, which is greater than 1.
Q13.
Answer :
xy2+xdx+y-x2ydy=0⇒xy2+1dx=yx2-1dy⇒xy2+1yx2-1=dydx⇒xy2+1dydx-yx2-1=0⇒y2+1dydx-yx-1x=0
In this differential equation, the order of the highest order derivative is 1 and its power is 1. So, it is a differential equation of degree 1 and order 1.
It is a non-linear equation, as the product containing dependent variable and its differential co-efficient y2dydx is present in it.
Q14.
Answer :
1-y2dx+1-x2dy=0⇒1-y2dx=-1-x2dy⇒-1-y21-x2=dydx⇒1-x2dydx+1-y2=0
In this differential equation, the order of the highest order derivative is 1 and its power is 1. So, it is a differential equation of order 1 and degree 1.
It is a non-linear equation, as the exponent of dependent variable y is more than 1 (on expanding 1-y2 binomially).
Q15.
Answer :
d2ydx2=dydx23Taking cubes of both sides, we get⇒d2ydx23=dydx2
In this differential equation, the order of the highest order derivative is 2 and its power is 3. So, it is a differential equation of order 2 and degree 3.
It is a non-linear differential equation, as it has degree 3, which is greater than 1.
Q16.
Answer :
2d2ydx2+31-dydx2-y=0⇒2d2ydx2=-31-dydx2-ySquaring both sides, we get⇒4d2ydx22=91-dydx2-y⇒4d2ydx22+9dydx2+9y-9=0
In this differential equation, the order of the highest order derivative is 2 and its power is 2. So, it is a differential equation of order 2 and degree 2.
It is a non-linear differential equation, as it has degree 2, which is greater than 1.
Q17.
Answer :
5d2ydx2=1+dydx232Squaring both sides, we get⇒25d2ydx22=1+dydx23⇒25d2ydx22=1+3dydx2+3dydx4+dydx6⇒25d2ydx22-dydx6-3dydx4-3dydx2-1=0
In this differential equation, the order of the highest order derivative is 2 and its power is 2. So, it is a differential equation of order 2 and degree 2.
It is a non-linear differential equation, as its degree is 2, which is greater than 1.
Q18.
Answer :
y=xdydx+a1+dydx2⇒y-xdydx=a1+dydx2Squaring both sides, we get⇒y-xdydx2=a21+dydx2⇒y2-2xydydx+x2dydx2=a2+a2dydx2⇒x2-a2dydx2-2xydydx+y2-a2=0
In this differential equation, the order of the highest order derivative is 1 and its highest power is 2. So, it is a differential equation of order 1 and degree 2.
It is a non-linear differential equation, as its degree is 2, which is greater than 1.
Q19.
Answer :
y=px+a2p2+b2⇒y-px=a2p2+b2Squaring both sides, we get⇒y-px2=a2p2+b2⇒y2-2pxy+p2x2=a2p2+b2⇒x2-a2p2-2pxy+y2-b2=0⇒x2-a2dydx2-2xydydx+y2-b2=0 Substituting p=dydx
In this differential equation, the order of the highest order derivative is 1 and its highest power is 2. So, it is a differential equation of order 1 and degree 2.
It is a non-linear differential equation, as its degree is 2, which is greater than 1.
Q20.
Answer :
d2ydx2+3dydx2=x2 log d2ydx2
In this differential equation, the order of the highest order derivative is 2.
Clearly, the R.H.S. of the differential equation cannot be expressed as a polynomial in d2ydx2.
Thus, its degree is not defined.
The order of the differential equation is 2 and its degree is not defined.
It is a non-linear differential equation, as one of its differential co-efficients, that is, dydx, has exponent 2, which is greater than 1.
Q21.
Answer :
d2ydx22+dydx2=x sin d2ydx2
In this differential equation, the order of the highest order derivative is 2.
Clearly, the R.H.S. of the differential equation cannot be expressed as a polynomial in d2ydx2. So, its degree is not defined.
The order of the differential equation is 2 and its degree is not defined.
It is a non-linear differential equation, as one of its differential co-efficients, that is, dydx, has exponent 2, which is more than 1.
Q22.
Answer :
y”2+y’3+sin y=0
In this differential equation, the order of the highest order derivative is 2 and its power is 2. So, the order of the differential equation is 2 and its degree is 2.
It is a non-linear differential equation, as its degree is 2, which is more than 1.
Q23.
Answer :
d2ydx2+5xdydx-6y=log x
In this differential equation, the order of the highest order derivative is 2 and its power is 1. So, the order of the differential equation is 2 and its degree is 1.
It is a linear differential equation.
Q24.
Answer :
d3ydx3+d2ydx2+dydx+y sin y=0
In this differential equation, the order of the highest order derivative is 3 and its power is 1. So, the order of the differential equation is 3 and its degree is 1.
It is a non-linear differential equation, as the exponent of the dependent variable is not equal to 1 (by expanding y.sin y).
Q25.
Answer :
dydx+ey=0
In this differential equation, the order of the highest order derivative is 1 and its power is 1. So, the order of the differential equation is 1 and its degree is 1.
It is a non-linear differential equation, as the exponent of the dependent variable is not equal to 1 (as per expansion series of ey).
Q26.
Answer :
dydx3-4dydx2+7y=sin x
In this differential equation, the order of the highest order derivative is 1 and its highest power is 3. So, the order of the differential equation is 1 and its degree is 3.
It is a non-linear differential equation, as its degree is 3, which is greater than 1.
Page 22.16 Ex. 22.2
Q1.
Answer :
The equation of the family of curves is
y2=x-c3 …(1)
where c∈R is a parameter.
This equation contains only one parameter, so we shall obtain a differential equation of first order.
Differentiating equation (1) with respect to x, we get
2ydydx=3x-c2 …(2)
Dividing equation (1) by equation (2), we get
y22ydydx=x-c33x-c2⇒y2dydx=x-c3⇒3y2dydx=x-c⇒c=x-3y2dydx
Substituting the value of c in equation (1), we get
y2=x-x+3y2dydx3⇒y2=27y38dydx3⇒8y2dydx3=27y3⇒8dydx3-27y=0
It is the required differential equation.
Q2.
Answer :
The equation of the family of curves is
y=emx …(1)
where m is a parameter.
This equation contains only one parameter, so we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
dydx=memx⇒dydx=my [Using equation (1)]
⇒m=1ydydx …(2)
Now, from equation (1), we get
ln y = ln emx⇒ln y= mx ln e⇒ln y= mx⇒m = 1xln y …(3)
Comparing equations (2) and (3), we get
1xln y = 1ydydx⇒xdydx=y ln y
It is the required differential equation.
Q3.
Answer :
(i) The equation of family of curves is
y2=4ax …(1)
where a is an arbitrary constant.
This equation contains only one arbitrary constant, so we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
2ydydx=4a⇒y2dydx=a 2
Putting the value of a in equation (1), we get
y2=4y2dydxx⇒y=2xdydx, It is the required differential equation.
(ii) The equation of family of curves is
y=cx+2c2+c3 …(1)
where c is an arbitrary constant.
This equation contains only one arbitrary constant, so we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
dydx=c …(2)
Putting the value of c in equation (1), we get
y=xdydx+2dydx2+dydx3
It is the required differential equation.
(iii) The equation of family of curves is
xy=a2 …(1)
where a is an arbitrary constant.
This equation contains only one arbitrary constant, so we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
y+xdydx=0
It is the required differential equation.
(iv) The equation of family of curves is
y=ax2+bx+c …(1)
where a, b and c are arbitrary constants. So, we shall get a differential equation of third order.
Differentiating equation (1) with respect to x, we get
dydx=2ax+b …(2)
Differentiating equation (2) with respect to x, we get
d2ydx2=2a …(3)
Differentiating equation (3) with respect to x, we get
d3ydx3=0
It is the required differential equation.
Q4.
Answer :
The equation of the family of curves is
y=Ae2x+Be-2x …(1)
where A and B are arbitrary constants.
This equation contains two arbitrary constants, so we shall get a differential equation of second order.
Differentiating equation (1) with respect to x, we get
dydx=2Ae2x-2Be-2x …(2)
Differentiating equation (2) with respect to x, we get
d2ydx2=4Ae2x+4Be-2x⇒d2ydx2=4Ae2x+Be-2x⇒d2ydx2=4yIt is the required differential equation.
Q5.
Answer :
The equation of the family of curves is
x=Acos nt+Bsin nt …(1)
where A and B are arbitrary constants.
This equation contains two arbitrary constants, so we shall get a differential equation of second order.
Differentiating equation (1) with respect to t, we get
dxdt=-Ansin nt+Bncos nt …(2)
Differentiating equation (2) with respect to t, we get
d2xdt2=-An2cos nt-Bn2sin nt⇒d2xdt2=-n2Acos nt+Bsin nt⇒d2xdt2=-n2x⇒d2xdt2+n2x=0 It is the required differential equation.
Q6.
Answer :
The equation of the family of curves is
y2=ab-x2, …(1)
where a and b are parameters.
This equation contains two arbitrary constants, so we shall get a differential equation of second order.
Differentiating equation (1) with respect to x, we get
2ydydx=-2ax …(2)
Differentiating equation (2) with respect to x, we get
dydx2+yd2ydx2=-a …(3)
From (2) and (3), we get
ydydx=xdydx2+yd2ydx2
It is the required differential equation.
Q7.
Answer :
The equation of the family of curves is
y2-2ay+x2=a2 …(1)
where a is a parameter.
This equation contains only one arbitrary constant, so we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
2ydydx-2adydx+2x=0⇒2ydydx+2x=2adydx⇒y+xdydx=a
Substituting the value of a in equation (2), we get
y2-2y+xdydxy+x2=y+xdydx2⇒y2dydx-2ydydx+xy+x2dydxdydx=ydydx+x2dydx2⇒y2dydx2-2y2dydx2-2xydydx+x2dydx2=y2dydx2+2xydydx+x2⇒x2-2y2dydx2-4xydydx-x2=0 It is the required differential equation.
Q8.
Answer :
The equation of the family of curves is
x-a2+y-b2=r2 …(1)
where a and b are parameters.
This equation contains two parameters, so we shall get a second order differential equation.
Differentiating equation (1) with respect to x, we get
2x-a+2y-bdydx=0 …(2)
Differentiating (2) with respect to x, we get
2+2dydx2+2y-bd2ydx2=0⇒1+dydx2+y-bd2ydx2=0⇒y-b=-1+dydx2d2ydx2 …(3)
From (2) and (3), we get
x-a-1+dydx2d2ydx2dydx=0⇒x-a=dydx+dydx3d2ydx2 …(4)
From (1), (3) and (4), we get
dydx+dydx32d2ydx22+1+dydx22d2ydx22=r2⇒dydx2+2dydx4+dydx6+1+2dydx2+dydx4d2ydx22=r2⇒dydx2+2dydx4+dydx6+1+2dydx2+dydx4=r2d2ydx22⇒1+3dydx2+3dydx4+dydx6=r2d2ydx22⇒1+dydx23=r2d2ydx22It is the required differential equation.
Q9.
Answer :
The equation of the family of circles that pass through the origin (0, 0) and whose centres lie on the y-axis is given by
x2+y-a2=a2 …(1)
where a is any arbitrary constant.
As this equation has only one arbitrary constant, we shall get a first order differential equation.
Differentiating equation (1) with respect to x, we get
2x+2y-adydx=0⇒x+y-adydx=0⇒x=a-ydydx⇒xdydx=a-y⇒a=y+xdydx …(2)
Substituting the value of a in equation (2), we get
x2+y-y-xdydx2=y+xdydx2⇒x2+x2dydx2=y2+2xydydx+x2dydx2⇒x2=y2+2xydydx⇒x2-y2dydx=2xy It is the required differential equation.
Q10.
Answer :
The equation of the family of circles that pass through the origin (0,0) and whose centres lie on the x-axis is given by
x-a2+y2=a2 …(1)
where a is any arbitrary constant.
As this equation has only one arbitrary constant, we shall get a first order differential equation.
Differentiating equation (1) with respect to x, we get
2x-a+2ydydx=0⇒x-a+ydydx=0⇒x+ydydx=a ..(2)
Substituting the value of a in equation (1), we get
x-x-ydydx2+y2=x+ydydx2⇒y2dydx2+y2=x2+2xydydx+y2dydx2⇒2xydydx+x2=y2 It is the required differential equation.
Q11.
Answer :
Let the surface area of the raindrop be A.
Thus, the rate of evaporation will be given by dVdt.
As per the given condition,
dVdt∝A⇒dVdt=-kA
Here, k is a constant. Also, the negative sign appears when V decreases and t increases.
Now,
V=43πr3
Here, r is the radius of the spherical drop.
∴ddt43πr3=-k×4πr2⇒43×3πr2drdt=-k×4πr2⇒drdt=-k It is the required differential equation.
Q12.
Answer :
The equation of the family of parabolas with latus rectum 4a and axis parallel to the x-axis is given by
y-β2=4ax-α …(1)
where α and β are two arbitrary constants.
As this equation has two arbitrary constants, we shall get second order differential equation.
Differentiating equation (1) with respect to x, we get
2y-βdydx=4a …(2)
Differentiating equation (2) with respect to x, we get
y-βd2ydx2+dydxdydx=0 …(3)
Now, from equation (2), we get
y-β=4adydx …(4)
From (3) and (4), we get
2adydxd2ydx2+dydx2=0⇒2ad2ydx2+dydx3=0 It is the required differential equation.
Q13.
Answer :
The given equation is
y=2×2-1+ce-x2 …(1)
where c is a parameter.
As this equation has one arbitrary constant, we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
dydx=22x+ce-x2(-2x)⇒dydx=4x-2xce-x2 …2
From (1) and (2), we get
dydx=4x-2xy-2×2+2
⇒dydx=4x-2xy+4×3-4x⇒dydx+2xy=4×3
Hence, y=2×2-1+ce-x2 is the solution to the differential equation dydx+2xy=4×3.
Q14.
Answer :
The equation of the family of non-vertical lines in a plane, say, the X‒Y plane, is
y=mx+c …(1)
where m and c are two arbitrary constants.
As this equation has two arbitrary constants, we shall get a second-order differential equation.
Differentiating (1) with respect to x, we get
dydx=m …(2)
Differentiating (2) with respect to x, we get
d2ydx2=0
It is the required differential equation.
Q15.
Answer :
(i) The equation of the family of curves is
2x+a2+y2=a2 …(1)
where a is a parameter.
As this equation has only one parameter, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
22x+a×2+2ydydx=0 …(2)
Now, from (1), we get
4×2+4ax+a2+y2=a2⇒4ax=-y2-4×2⇒a=-4×2+y24x
Putting the value of a in (2), we get
42x-4×2+y24x+2ydydx=0⇒48×2-4×2-y24x+2ydydx=0⇒4×2-y2+2xydydx=0⇒y2-4×2-2xydydx=0It is the required differential equation.
(ii) The equation of the family of curves is
2x-a2-y2=a2⇒4×2-4ax+a2-y2=a2⇒4×2-4ax-y2=0 …1
where a is a parameter.
As this equation has only one parameter, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
8x-4a-2ydydx=0⇒-ydydx+4x=2a …2
Now, from (1), we get
2a=4×2-y22x …(3)
From (2) and (3), we get
-ydydx+4x=4×2-y22x⇒-2xydydx+8×2=4×2-y2⇒-2xydydx+4×2+y2=0⇒2xydydx=4×2+y2It is the required differential equation.
(iii) The equation of the family of curves is
x-a2+2y2=a2⇒x2-2ax+a2+2y2=a2⇒x2-2ax+2y2=0 …1
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
2x-2a+4ydydx=0 …(2)
Now, from (1), we get
2a=x2+2y2x …(3)
From (2) and (3), we get
2x-x2+2y2x+4ydydx=0⇒2×2-x2-2y2+4xydydx=0⇒4xydydx+x2-2y2=0⇒4xydydx=2y2-x2It is the required differential equation.
Q16.
Answer :
(i) The equation of the family of curves is
x2+y2=a2 …(1)
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
2x+2ydydx=0⇒x+ydydx=0It is the required differential equation.
(ii) The equation of family of curves is
x2-y2=a2 …(1)
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
2x-2ydydx=0⇒x-ydydx=0It is the required differential equation.
(iii) The equation of family of curves is
y2=4ax …(1)
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
2ydydx=4a⇒2ydydx=y2x Using 1⇒2xdydx=y⇒y-2xdydx=0It is the required differential equation.
(iv) The equation of family of curves is
x2+y-b2=1 …(1)
where b is a parameter.
As this equation contains only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
2x+2y-bdydx=0⇒2x+21-x2dydx=0 Using 1⇒x=-1-x2dydx⇒x2=1-x2dydx2⇒x2=dydx2-x2dydx2⇒x21+dydx2=dydx2It is the required differential equation.
(v) The equation of family of curves is
x-a2-y2=1 …(1)
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
2x-a-2ydydx=0⇒x-a-ydydx=0⇒1+y2=ydydx Using 1⇒1+y2=y2dydx2⇒y2dydx2-y2=1It is the required differential equation.
(vi) The equation of family of curves is
x2a2-y2b2=1 …(1)
where a and b are parameters.
As this equation has two arbitrary constants, we shall get a differential equation of second order.
Differentiating (1) with respect to x, we get
2xa2-2yb2dydx=0, …(2)
Differentiating (2) with respect to x, we get
2a2-2b2dydx2-2yb2d2ydx2=0⇒2a2=2b2yd2ydx2+dydx2⇒b2a2=yd2ydx2+dydx2 …3
Now, from (2), we get
2xa2=2yb2dydx⇒b2a2=yxdydx …4
From (3) and (4), we get
yxdydx=yd2ydx2+dydx2⇒xyd2ydx2+dydx2=ydydxIt is the required differential equation.
(vii) The equation of family of curves is
y2=4ax-b …(1)
where a and b are parameters.
As this equation has two arbitrary constants, we shall get a differential equation of second order.
Differentiating (1) with respect to x, we get
2ydydx=4a⇒ydydx=2a …2
Differentiating (2) with respect to x, we get
yd2ydx2+dydx2=0
It is the required differential equation.
(viii) The equation of family of curves is
y=ax3 …(1)
where a is a parameter.
As this equation has only one arbitrary constant, so we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
dydx=3ax2⇒dydx=3×yx3×x2 Using 1⇒xdydx=3yIt is the required differential equation.
(ix) The equation of family of curves is
x2+y2=ax3 …(1)
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
2x+2ydydx=3ax2⇒2x+2ydydx=3×2+y2x3x2 Using 1⇒2x+2ydydx=3×2+y2x⇒2×2+2xydydx=3×2+3y2⇒2xydydx=x2+3y2It is the required differential equation.
(x) The equation of family of curves is
y=eax⇒log y=ax …1
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
1ydydx=a⇒1ydydx=log yx Using 1⇒xdydx=y log yIt is the required differential equation.
Q17.
Answer :
The equation of the family of ellipses having centre at the origin and foci on the x-axis is
x2a2+y2b2= 1 …(1)
where a and b are the parameters.
As this equation contains two parameters, we shall get a second-order differential equation.
Differentiating (1) with respect to x, we get
2xa2+2yb2dydx=0 …(2)
Differentiating (2) with respect to x, we get
2a2+2b2dydx2+yd2ydx2=0⇒2a2=-2b2dydx2+yd2ydx2⇒b2a2=-dydx2+yd2ydx2 …(3)
Now, from (2), we get
xa2=-yb2dydx⇒b2a2=-yxdydx …(4)
From (3) and (4), we get
-yxdydx=-dydx2+yd2ydx2⇒yxdydx=dydx2+yd2ydx2⇒ydydx=xdydx2+xyd2ydx2⇒xyd2ydx2+xdydx2-ydydx=0 It is the required differential equation.
Q18.
Answer :
The equation of the family of hyperbolas having the centre at the origin and foci on the x-axis is
x2a2-y2b2=1 …(1)
where a and b are parameters.
As this equation contains two parameters, we shall get a second-order differential equation.
Differentiating equation (1) with respect to x, we get
2xa2-2yb2dydx=0 …(2)
Differentiating equation (2) with respect to x, we get
2a2-2b2yd2ydx2+dydx2=0⇒1a2=1b2yd2ydx2+dydx2⇒b2a2=yd2ydx2+dydx2 3
Now, from equation (2), we get
2xa2=2yb2dydx⇒b2a2=yxdydx …(4)
From (3) and (4), we get
yxdydx=yd2ydx2+dydx2⇒ydydx=xyd2ydx2+xdydx2⇒xyd2ydx2+xdydx2-ydydx=0 It is the required differential equation.
Q19.
Answer :
The equation of the family of circles in the second quadrant and touching the co-ordinate axes is
x+a2+y-a2=a2⇒x2+2ax+a2+y2-2ay+a2=a2⇒x2+2ax+y2-2ay+a2=0 …(1)
where a is a parameter.
As this equation contains one parameter, we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
2x+2a+2ydydx-2adydx=0⇒x+ydydx+a-adydx=0⇒x+ydydx+a1-dydx=0⇒a=x+ydydxdydx-1 … (2)
From (1) and (2), we get
x2+2xx+ydydxdydx-1+y2-2yx+ydydxdydx-1+x+ydydxdydx-12=0⇒x2dydx-12+2xx+ydydxdydx-1+y2dydx-12-2yx+ydydxdydx-1+x+ydydx2=0⇒x2dydx2-2x2dydx+x2+2xxdydx-x+ydydx2-ydydx+y2dydx2-2dydx+1-2yxdydx-x+ydydx2-ydydx+x2+2xydydx+y2dydx2=0⇒x2+2xydydx+y2dydx2=x2+2xy+y2+x2+2xy+y2dydx2⇒x+ydydx2=x+y21+dydx2 It is the required differential equation.
Page 22.24 Ex. 22.3
Q1.
Answer :
We have,
y=bex+ce2x …(1)
Differentiating both sides of equation (1) with respect to x, we get
dydx=bex+2ce2x …(2)
Differentiating both sides of equation (2) with respect to x, we get
d2ydx2=bex+4ce2x= 3bex+6ce2x-2bex-2ce2x= 3bex+2ce2x-2bex+ce2x= 3dydx-2y Using equations 1 and 2
⇒d2ydx2-3dydx+2y=0
Hence, the given function is the solution to the given differential equation.
Q2.
Answer :
We have,
y=4 sin 3x …(1)
Differentiating both sides of equation (1) with respect to x, we get
dydx=12 cos 3x …(2)
Differentiating both sides of equation (2) with respect to x, we get
d2ydx2=-36 sin 3x⇒d2ydx2=-94 sin 3x⇒d2ydx2=-9y Using equation 1
⇒d2ydx2+9y=0
Hence, the given function is the solution to the given differential equation.
Q3.
Answer :
We have,
y=ae2x+be-x …(1)
Differentiating both sides of equation (1) with respect to x, we get
dydx=2ae2x-be-x …(2)
Differentiating both sides of equation (2) with respect to x, we get
d2ydx2=4ae2x+be-x⇒d2ydx2=2ae2x-be-x+2ae2x+2be-x⇒d2ydx2=2ae2x-be-x+2ae2x+be-x⇒d2ydx2=dydx+2y Using equations 1 and 2
⇒d2ydx2-dydx-2y=0
Hence, the given function is the solution to the given differential equation.
Q4.
Answer :
We have,
y=A cos x+B sin x …(1)
Differentiating both sides of equation (1) with respect to x, we get
dydx=-A sin x+B cos x …(2)
Differentiating both sides of equation (2) with respect to x, we get
d2ydx2=-A cos x-B sin x⇒d2ydx2=-A cos x+B sin x⇒d2ydx2=-y Using equation 1
⇒d2ydx2+y=0
Hence, the given function is the solution to the given differential equation.
Q5.
Answer :
We have,
y=A cos 2x-B sin 2x …(1)
Differentiating both sides of (1) with respect to x, we get
dydx=-2A sin 2x-2B cos 2x …(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=-4A cos 2x+4B sin 2x⇒d2ydx2=-4A cos 2x-B sin 2x⇒d2ydx2=-4y Using 1
⇒d2ydx2+4y=0
Hence, the given function is the solution to the given differential equation.
Q6.
Answer :
We have,
y=AeBx …(1)
Differentiating both sides of (1) with respect to x, we get
dydx=ABeBx …(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=AB2eBx⇒d2ydx2=ABeBx2AeBx⇒d2ydx2=1ydydx2 Using 1 and 2⇒d2ydx2=1ydydx2
Hence, the given function is the solution to the given differential equation.
Q7.
Answer :
We have,
y=ax+b …(1)
Differentiating both sides of (1) with respect to x, we get
dydx=-ax2 …(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=2ax3⇒d2ydx2=-2x-ax2⇒d2ydx2=-2xdydx Using 2⇒d2ydx2+2xdydx=0
Hence, the given function is the solution to the given differential equation.
Q8.
Answer :
We have,
y2=4ax …(1)
Differentiating both sides of (1) with respect to x, we get
2ydydx=4a
⇒dydx=2ay …(2)
Now, differentiating both sides of (1) with respect to y, we get
2y=4adxdy
⇒dxdy=y2a …(3)
∴ xdydx+adxdy=x2ay+ay2a Using 2 and 3⇒xdydx+adxdy=2axy+y2⇒xdydx+adxdy= y22y+y2 Using 1⇒xdydx+adxdy= y2+y2⇒xdydx+adxdy=y⇒y=xdydx+adxdy
Hence, the given function is the solution to the given differential equation.
Q9.
Answer :
We have,
Ax2+By2=1 …(1)
Differentiating both sides of (1) with respect to x, we get
2Ax+2Bydydx=0 …(2)
Differentiating both sides of (2) with respect to x, we get
2A+2Bdydx2+2Byd2ydx2=0⇒2Byd2ydx2+dydx2=-2A⇒ydydx+dydx2=-2A2B⇒ydydx+dydx2=–yxdydx Using 2⇒xydydx+dydx2=ydydx
Hence, the given function is the solution to the given differential equation.
Q10.
Answer :
We have,
y=ax3+bx2+c …(1)
Differentiating both sides of (1) with respect to x, we get
dydx=3ax2+2bx …(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=6ax+2b …(3)
Differentiating both sides of (3) with respect to x, we get
d3ydx3=6a
Hence, the given function is the solution to the given differential equation.
Q11.
Answer :
We have,
y=c-x1+cx …(1)
Differentiating both sides of (1) with respect to x, we get
dydx=1+cx-1-c-xc1+cx2⇒dydx=-1-cx-c2+cx1+cx2⇒dydx=-1+c21+cx2 …2
Now,
1+x2dydx+1+y2=-1+x21+c21+cx2+1+c-x21+cx2 Using 1 and 2=-1+x21+c21+cx2+1+cx2+c-x21+cx2=-1+x21+c21+cx2+1+2cx+c2x2+c2-2cx+x21+cx2=-1+x21+c21+cx2+1+x2+c21+x21+cx2=-1+x21+c21+cx2+1+x21+c21+cx2=0⇒1+x2dydx+1+y2=0
Hence, the given function is the solution to the given differential equation.
Q12.
Answer :
We have,
y=exA cos x+B sin x …(1)
Differentiating both sides of (1) with respect to x, we get
dydx=exA cos x+B sin x+ex-A sin x+B cos x …(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=exA cos x+B sin x+ex-A sin x+B cos x+ex-A sin x+B cos x+ex-A cos x-B sin x⇒d2ydx2=2ex-A sin x+B cos x⇒d2ydx2=2ex-A sin x+B cos x+2exA cos x+B sin x-2exA cos x+B sin x⇒d2ydx2=2dydx-2y Using 1 and 2⇒d2ydx2-2dydx+2y=0
Hence, the given function is the solution to the given differential equation.
Q13.
Answer :
We have,
y=cx+2c2 …(1)
Differentiating both sides of (1) with respect to x, we get
dydx=c …(2)
Now,
2dydx2+xdydx-y=2c2+cx-cx-2c2=0 Using 1 and 2⇒2dydx2+xdydx-y=0
Hence, the given function is the solution to the given differential equation.
Q14.
Answer :
We have,
y=-x-1 …(1)
Differentiating both sides of (1) with respect to x, we get
dydx=-1 …(2)
Now,
dydx-y2-x2y-x=dydx-y+x=-1–x-1+x Using 1 and 2=-1+1=0⇒dydx=y2-x2y-x⇒y-xdy=y2-x2dx⇒y-xdy-y2-x2dx=0
Hence, the given function is the solution to the given differential equation.
Q15.
Answer :
We have,
y2=4ax+a …(1)
Differentiating both sides of (1) with respect to x, we get
2ydydx=4a⇒ydydx=2a⇒dydx=2ay …2
Now,
y1-dydx2-2xdydx=y1-4a2y2-2x2ay=yy2-4a2y2-4axy=y2-4a2y-4axy=4ax+4a2-4a2y-4axy Using 1=4axy-4axy=0⇒y1-dydx2=2xdydx
Hence, the given function is the solution to the given differential equation.
Q16.
Answer :
We have,
y=cetan-1x …(1)
Differentiating both sides of (1) with respect to x, we get
dydx=cetan-1×11+x2 …(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=c1+x2etan-1×11+x2-etan-1x2x1+x22⇒d2ydx2=cetan-1x-2xetan-1×1+x22⇒d2ydx2=c1-2xetan-1×1+x22⇒1+x2d2ydx2=c1-2xetan-1×1+x2⇒1+x2d2ydx2=1-2xdydx Using 2⇒1+x2d2ydx2+2x-1dydx=0
Hence, the given function is the solution to the given differential equation.
Q17.
Answer :
We have,
y=em cos-1x …(1)
Differentiating both sides of (1) with respect to x, we get
dydx=mem cos-1x-11-x2⇒dydx=-mem cos-1×1-x2 …2
Differentiating both sides of (2) with respect to x, we get
d2ydx2=ddx-mem cos-1×1-x2⇒d2ydx2=-m1-x2mem cos-1x-11-x2-em cos-1×12-2×1-x21-x2⇒1-x2d2ydx2=-m-mem cos-1x+xem cos-1×1-x2⇒1-x2d2ydx2=m2em cos-1x-mxem cos-1×1-x2⇒1-x2d2ydx2=m2y+xdydx Using 1 and 2⇒1-x2d2ydx2-xdydx-m2y=0
Hence, the given function is the solution to the given differential equation.
Page 22.25 Ex. 22.3
Q18.
Answer :
We have,
y=log x+x2+a22 …(1)
Differentiating both sides of (1) with respect to x, we get
dydx=ddxlog x+x2+a22=ddx2 log x+x2+a2=21+122xx2+a2x+x2+a2=2×2+a2+xx2+a2x+x2+a2=2×2+a2 …2
Differentiating both sides of (2) with respect to x, we get
d2ydx2=2-122xx2+a2x2+a2⇒a2+x2d2ydx2=-2xx2+a2⇒a2+x2d2ydx2=-xdydx Using 2⇒a2+x2d2ydx2+xdydx=0
Hence, the given function is the solution to the given differential equation.
Q19.
Answer :
We have,
y=2×2-1+ce-x2 …(1)
Differentiating both sides of (1) with respect to x, we get
dydx=4x-ce-x22x=2×2-ce-x2=-2x2x2-2+ce-x2-2×2=-2x2x2-1+ce-x2-2×2=-2xy-2×2 Using 1⇒dydx=-2xy+4×3⇒dydx+2xy=4×3
Hence, the given function is the solution to the given differential equation.
Q20.
Answer :
We have,
y=e-x+ax+b … (i)
Differentiating both sides of equation (i) with respect to x, we have
dydx=-e-x+a … (ii)
Differentiating both sides of equation (ii) with respect to x, we have
d2ydx2=e-x⇒exd2ydx2=1
Hence, the given function is a solution of the given differential equation.
Q21.
Answer :
(i) We have,
y=ax …..1
Given differential equation: xdydx=y
Differentiating both sides of (1) with respect to x, we get
dydx=a⇒dydx=yx Using 1⇒xdydx=y
Hence, the given function is the solution to the given differential equation.
(ii) We have,
y=±a2-x2⇒y2=a2-x2 …..1
Given differential equation: x+ydydx=0
Differentiating both sides of (1) with respect to x, we get
2y dydx=-2x⇒y dydx=-x⇒x+y dydx=0
Hence, the given function is the solution to the given differential equation.
(iii) We have,
y=ax+a⇒xy+ay=a⇒xy=a1-y⇒xy1-y=a⇒1-yxy=1a …..1
given differential equation: xdydx+y=y2
Differentiating both sides of (1) with respect to x, we get
xy0-dydx-1-yxdydx+yxy2=0⇒xy-dydx-1-yxdydx+y=0⇒-xydydx-xdydx-y+xydydx+y2=0⇒-xdydx-y+y2=0⇒xdydx+y=y2
Hence, the given function is the solution to the given differential equation.
(iv) We have,
y=ax+b+12x …..1
Differentiating both sides of (1) with respect to x, we get
dydx=a-12×2 …..2Now differentiating both sides of 2 with respect to x, we get⇒d2ydx2=-12×-2×3⇒d2ydx2=1×3⇒x3 d2ydx2=1
Hence, the given function is the solution to the given differential equation.
(v) We have,
y=14x±a2 …..1
Differentiating both sides of (1) with respect to x, we get
dydx=14×2x±a⇒dydx=12x±aSquaring both sides we get⇒dydx2=12x±a2⇒dydx2=14x±a2⇒dydx2=y Using 1∴y=dydx2
Hence, the given function is the solution to the given differential equation.
Page 22.27 Ex. 22.4
Q1.
Answer :
We have,y=log x …(1)Differentiating both sides of (1) with respect to x, we getdydx=1xor, xdydx=1It is the given differential equation.Thus, y=log x satisfies the given differential equation.Hence, it is a solution.Also, when x=1, y=log 1=0, i.e., y1=0.Hence, y=log x is the solution to the given initial value problem.
Q2.
Answer :
We have,
y=ex …(1)
Differentiating both sides of (1) with respect to x, we get
dydx=ex
⇒dydx=y [Using (1)]
It is the given differential equation.
Here, y=ex satisfies the given differential equation; hence, it is a solution.
Also, when x=0, y=e0=1, i.e., y(0)=1.
Hence, y=ex is the solution to the given initial value problem.
Q3.
Answer :
We have,
y=sin x …(1)
Differentiating both sides of (1) with respect to x, we get
dydx=cos x …(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=-sin x⇒d2ydx2=-y Using 1
⇒d2ydx2+y=0
It is the given differential equation.
Here, y=sin x satisfies the given differential equation; hence, it is a solution.
Also, when x=0, y=sin 0=0, i.e., y0=0.
And, when x=0, y’=cos 0=1, i.e., y’0=1.
Hence, y=sin x is the solution to the given initial value problem.
Q4.
Answer :
We have,
y=ex+1 …(1)
Differentiating both sides of (1) with respect to x, we get
dydx=ex …(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=ex⇒d2ydx2=dydx Using 2⇒d2ydx2-dydx=0 It is the given differential equation.
y=ex+1 satisfies the given differential equation; hence, it is a solution.
Also, when x=0, y=e0+1=1+1=2, i.e. y0=2.
And, when x=0, y’=e0=1, i.e. y’0=1.
Hence, y=ex+1 is the solution to the given initial value problem.
Page 22.28 Ex. 22.4
Q5.
Answer :
We have,
y=e-x+2 …(1)
Differentiating both sides of (1) with respect to x, we get
dydx=-e-x⇒dydx=-y-2 Using 1⇒dydx+y=2 It is the given differential equation.
y=e-x+2 satisfies the given differential equation; hence, it is a solution.
Also, when x=0, y=e0+2=1+2=3, i.e. y0=3.
Hence, y=e-x+2 is the solution to the given initial value problem.
Q6.
Answer :
We have,
y=sin x+ cos x …..(1)
Differentiating both sides of (1) with respect to x, we get
dydx=cos x-sin x …..(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=-sin x-cos x⇒d2ydx2=-sin x+cos x⇒d2ydx2=-y Using 1
⇒d2ydx2+y=0
It is the given differential equation.
Therefore, y=sin x+ cos x satisfies the given differential equation.
Also, when x=0; y=sin 0+cos 0=1, i.e. y0=1.
And, when x=0; y’=cos 0-sin 0=1, i.e. y’0=1.
Hence, y=sin x+ cos x is the solution to the given initial value problem.
Q7.
Answer :
We have,
y = ex + e-x …..(1)
Differentiating both sides of (1) with respect to x, we get
dydx=ex-e-x …..(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=ex+e-x⇒d2ydx2=y Using 1
⇒d2ydx2-y=0
It is the given differential equation.
Therefore, y = ex + e-x satisfies the given differential equation.
Also, when x=0; y=e0+e0=1+1, i.e. y0=2.
And, when x=0; y1=e0-e0=1-1, i.e. y’0=0.
Hence, y = ex + e-x is the solution to the given initial value problem.
Q8.
Answer :
We have,
y = ex + e2x …..(1)
Differentiating both sides of (1) with respect to x, we get
dydx=ex+2e2x …..(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=ex+4e2x⇒d2ydx2=3ex+2e2x-2ex+e2x⇒d2ydx2=3dydx-2y Using 1 and 2⇒d2ydx2-3dydx+2y=0
⇒d2ydx2-3dydx+2y=0
It is the given differential equation.
Therefore, y = ex + e2x satisfies the given differential equation.
Also, when x=0; y=e0+e0=1+1, i.e. y0=2.
And, when x=0; y’=e0+2e0=1+2, i.e. y’0=3.
Hence, y = ex + e2x is the solution to the given initial value problem.
Disclaimer: In the question instead of y(0) = 1, it should have been y(0) = 2.
Q9.
Answer :
We have,
y = xex + ex …..(1)
Differentiating both sides of (1) with respect to x, we get
dydx=xex+ex+ex⇒dydx=xex+2ex ….2
Differentiating both sides of (2) with respect to x, we get
d2ydx2=xex+ex+2ex⇒d2ydx2=xex+3ex⇒d2ydx2=2xex+2ex-xex+ex⇒d2ydx2=2dydx-y Using 1 and 2⇒d2ydx2-2dydx+y=0
d2ydx2-2dydx+y=0
It is the given differential equation.
Thus, y = xex + ex satisfies the given differential equation.
Also, when x=0, y=0+1=1, i.e. y0=1.
And, when x=0, y’=0+2=2, i.e. y’0=2.
Hence, y = xex + ex is the solution to the given initial value problem.
Page 22.33 Ex. 22.5
Q1.
Answer :
We have, dydx=x2+x-1x⇒dy=x2+x-1xdxIntegrating both sides, we get⇒∫dy=∫x2+x-1xdx⇒y=x33+x22-logx+CClearly, y=x33+x22-logx+C is defined for all x∈R except x=0.Hence, y=x33+x22-logx+C, where x∈R-0, is the solution to the given differential equation.
Q2.
Answer :
We have, dydx=x5+x2-2x⇒dy=x5+x2-2xdxIntegrating both sides, we get⇒∫dy=∫x5+x2-2xdx⇒y=x66+x33-2logx+CClearly, y=x66+x33-2logx+C is defined for all x∈R except x=0.Hence, y=x66+x33-2logx+C, where x∈R-0, is the solution to the given differential equation.
Q3.
Answer :
We have, dydx+2x=e3x⇒dydx=e3x-2x⇒dy=e3x-2xdxIntegrating both sides, we get⇒∫dy=∫e3x-2xdx⇒y=e3x3-2×22+C⇒y=e3x3-x2+C⇒y+x2=e3x3+CSo, y+x2=e3x3+C is defined for all x∈R. Hence, y+x2=e3x3+C, where x∈R, is the solution to the given differential equation.
Q4.
Answer :
We have, x2+1dydx=1⇒dydx=1×2+1⇒dy=1×2+1dxIntegrating both sides, we get⇒∫dy=∫1×2+1dx⇒y=tan-1x+CSo, y=tan-1x+C is defined for all x∈R. Hence, y=tan-1x+C, where x∈R, is the solution to the given differential equation.
Q5.
Answer :
We have, dydx=1-cos x1+cos x⇒dydx=2sin2 x22cos2 x2⇒dydx=tan2 x2⇒dy=tan2 x2dxIntegrating both sides, we get∫dy=∫tan2 x2dx⇒∫dy=∫sec2 x2-1dx⇒y= 2 tan x2-x+CSo, y= 2 tan x2-x+C is defined for all x∈R. Hence, y= 2 tan x2-x+C, where x∈R, is the solution to the given differential equation.
Q6.
Answer :
We have, x+2dydx=x2+3x+7⇒dydx=x2+3x+7x+2⇒dy=x2+3x+7x+2dxIntegrating both sides, we get∫dy=∫x2+3x+7x+2dx⇒∫dy=∫x2+3x+2+5x+2dx⇒∫dy=∫x+2x+1+5x+2dx⇒∫dy=∫x+1+5x+2dx⇒y=x22+x+5 logx+2+CSo, y=x22+x+5 logx+2+C is defined for all x∈R except x=-2.Hence, y=x22+x+5 logx+2+C, where x∈R-2, is the solution to the given differential equation.
Q7.
Answer :
We have, dydx=tan-1x⇒dy=tan-1xdxIntegrating both sides, we get∫dy=∫tan-1xdx⇒y=∫1II×tan-1xI dx ⇒y=tan-1x∫1 dx-∫ddxtan-1x∫1 dxdx⇒y=x tan-1x-∫x1+x2dx⇒y=x tan-1x-12∫2×1+x2dx⇒y=x tan-1x-12log1+x2+CSo, y=x tan-1x-12log1+x2+C is defined for all x∈R. Hence, y=x tan-1x-12log1+x2+C is the solution to the given differential equation.
Q8.
Answer :
We have, dydx=log x⇒dy=log xdxIntegrating both sides, we get∫dy=∫log xdx⇒∫dy=∫1II×log xI dx ⇒∫dy=log x∫1 dx-∫ddxlog x∫1 dxdx⇒y=xlog x-∫xxdx⇒y=xlog x-∫1dx⇒y=xlog x-x⇒y=xlog x-1+C⇒y=xlog x-1+CSo, y=xlog x-1+C is defined for all x∈R except x=0. Hence, y=xlog x-1+C, where x∈R-0, is the solution to the given differential equation.
Q9.
Answer :
We have, 1xdydx=tan-1 x⇒dydx=x tan-1x⇒dy=x tan-1xdxIntegrating both sides, we get∫dy=∫x tan-1xdx⇒y=∫xII×tan-1xI dx ⇒∫dy=tan-1x∫x dx-∫ddxtan-1x∫x dxdx⇒y=x2 tan-1×2-12∫x21+x2dx⇒y=x2 tan-1×2-12∫x2+1-11+x2dx⇒y=x2 tan-1×2-12∫1-11+x2dx⇒y=x2 tan-1×2-12x+tan-1×2+C⇒y=x2+1 tan-1×2-12x+CHence, y=x2+1 tan-1×2-12x+C is the solution to the given differential equation.
Q10.
Answer :
We have, dydx=cos3 x sin2 x+x2x+1⇒dy=cos3 x sin2 x+x2x+1dxIntegrating both sides, we get∫dy=∫cos3 x sin2 x+x2x+1dx⇒y=∫cos3 x sin2 x dx +∫x2x+1dx ⇒y=I1 +I2 …..1
where I1=∫cos3 x sin2 x dx I2=∫x2x+1dxNow, I1=∫cos3 x sin2 x dx=∫sin2 x 1-sin2 xcos x dxPutting t= sin x, we getdt=cos x dx⇒ I1=∫t2 1-t2dt=∫t2-t4dt=t33-t55+C1=sin3 x3-sin5 x5+C1 I2=∫x2x+1dx
Putting t2=2x+1, we get2t dt=2dx⇒tdt=dxNow, I2=∫t2-12t× t dt=12∫t4-t2 dt=t510-t36+C2=2x+15210-2x+1326+C2
Putting the values of I1 and I2 in1, we gety=sin3 x3-sin5 x5+C1+2x+15210-2x+1326+C2 y=sin3 x3-sin5 x5+2x+15210-2x+1326+C Where, C=C1+C2Hence, y=sin3 x3-sin5 x5+2x+15210-2x+1326+C is the solution to the given differential equation.
Q11.
Answer :
We have, dydx=cos3 x sin2 x+x2x+1⇒dy=cos3 x sin2 x+x2x+1dxIntegrating both sides, we get∫dy=∫cos3 x sin2 x+x2x+1dx⇒y=∫cos3 x sin2 x dx +∫x2x+1dx ⇒y=I1 +I2 …..1
where I1=∫cos3 x sin2 x dx I2=∫x2x+1dxNow, I1=∫cos3 x sin2 x dx=∫sin2 x 1-sin2 xcos x dxPutting t= sin x, we getdt=cos x dx⇒ I1=∫t2 1-t2dt=∫t2-t4dt=t33-t55+C1=sin3 x3-sin5 x5+C1 I2=∫x2x+1dx
Putting t2=2x+1, we get2t dt=2dx⇒tdt=dxNow, I2=∫t2-12t× t dt=12∫t4-t2 dt=t510-t36+C2=2x+15210-2x+1326+C2
Putting the values of I1 and I2 in1, we gety=sin3 x3-sin5 x5+C1+2x+15210-2x+1326+C2 y=sin3 x3-sin5 x5+2x+15210-2x+1326+C Where, C=C1+C2Hence, y=sin3 x3-sin5 x5+2x+15210-2x+1326+C is the solution to the given differential equation.
Q12.
Answer :
We have, dydx-x sin2 x=1xlog x⇒dydx=1xlog x+x sin2 x⇒dydx=1xlog x+x 21-cos 2x⇒dydx=1xlog x+x 2-x 2cos 2x⇒dy=1xlog x+x 2-x 2cos 2xdxIntegrating both sides, we get∫dy=∫1xlog x+x 2-x 2cos 2xdx⇒y=∫1xlog xdx+12∫x dx-12∫x cos 2xdx⇒y=loglog x+12×x22-12∫xI×cos 2xII dx ⇒y=loglog x+x24-x2∫cos 2xdx+12∫ddxx∫cos 2x dxdx⇒y=loglog x+x24-xsin 2×4-cos 2×8+CHence, y=loglog x+x24-xsin 2×4-cos 2×8+C is the solution to the given differential equation.
Q13.
Answer :
We have,dydx=x5 tan-1×3⇒dy=x5 tan-1x3dxIntegrating both sides, we get∫dy=∫x5 tan-1x3dx⇒y=∫x5 tan-1x3dxPutting t=x3, we getdt=3x2dx∴ y=13∫t tan-1 t dt=13∫tII×tan-1 tI dx =13tan-1 t∫t dt-∫ddttan-1 t∫t dxdt=13×t2 tan-1 t2-16∫t21+t2dt=t2 tan-1 t6-16∫t2+1-11+t2dt=t2 tan-1 t6-16∫dt+16∫11+t2dt=t2 tan-1 t6-16t+tan-1 t6 +C=x6 tan-1 x36-16×3+tan-1 x36+C=16×6 tan-1 x3-x3+tan-1 x3+CHence, y=16×6 tan-1 x3-x3+tan-1 x3+C is the solution to the given differential equation.
Q14.
Answer :
We have, sin4 xdydx= cos x⇒dy=cos xsin4 xdxIntegrating both sides, we get⇒∫dy=∫cos xsin4 xdx⇒y=∫cos xsin4 xdxPutting sin x=t⇒cos x dx=dt∴ y=∫1t4dt=t-3-3+C=-sin-3 x3+C=-13cosec3 x+C Hence, y=-13cosec3 x+C is the solution to the given differential equation.
Q15.
Answer :
We have, cos xdydx-cos 2x=cos 3x⇒dy=cos 3x+cos 2xcos xdx⇒dy=4cos3 x-3cos x+2cos2 x-1cos xdx⇒dy=4cos2 x-3+2cos x-sec xdx⇒dy=22cos2 x-1-1+2cos x-sec xdx⇒dy=2cos 2x-1+2cos x-sec xdxIntegrating both sides, we get∫dy=∫2cos 2x-1+2cos x-sec xdx⇒y=sin 2x-x+2sin x-logsec x+tan x+CHence, y=sin 2x-x+2sin x-logsec x+tan x+C is the solution to the given differential equation.
Q16.
Answer :
We have, 1-x4dy=x dx⇒dy=x1-x4dxIntegrating both sides, we get∫dy=∫x1-x4dx⇒y=∫x1-x4dxPutting x2=t⇒2x dx=dt∴y=12∫dt1-t2=sin-1t2+C=12sin-1×2+CHence, y=12sin-1×2+C is the solution to the given differential equation.
Q17.
Answer :
We have, a+xdy+xdx=0⇒a+xdy=-xdx⇒dy=-xa+xdx⇒dy=-x+a-aa+xdx⇒dy=-a+x-aa+xdxIntegrating both sides, we get∫dy=-∫a+x-aa+xdxdx⇒y=-2a+x323+2aa+x+C⇒y+23a+x32-2aa+x=CHence, y+23a+x32-2aa+x=C is the solution to the given differential equation.
Q18.
Answer :
We have,1+x2dydx-x=2tan-1x⇒1+x2dydx=x+2tan-1x⇒dy=x1+x2+21+x2tan-1xdx⇒dy=12×2×1+x2+21+x2tan-1x dxIntegrating both sides, we get∫dy=∫12×2×1+x2+21+x2tan-1x dx⇒y=12∫2×1+x2dx+2∫11+x2tan-1x dx⇒y=12log1+x2+2∫11+x2tan-1x dxPutting tan-1x=t⇒11+x2dx=dt∴y=12log1+x2+2∫t dt=12log1+x2+t2+C=12log1+x2+tan-1×2+CHence, y=12log1+x2+tan-1×2+C is the solution to the given differential equation.
Q19.
Answer :
We have, dydx=x log x⇒dy=x log xdxIntegrating both sides, we get∫dy=∫x log xdx⇒y=∫xII×log xI dx ⇒y=log x∫x dx-∫ddxlog x∫x dxdx⇒y=log x×x2 2-∫1x×x22dx⇒y=12×2 log x-∫x2dx⇒y=12×2 log x-x24+CHence, y=12×2 log x-x24+C is the solution to the given differential equation.
Q20.
Answer :
We have, dydx=x ex-52+cos2 x⇒dydx=x ex-52+cos 2×2+12⇒dydx=x ex+cos 2×2-2⇒dy=x ex+cos 2×2-2dxIntegrating both sides, we get∫dy=∫x ex+cos 2×2-2dx⇒y=∫x ex dx+12∫cos 2x dx-2∫dx⇒y=x∫ex dx-∫ddxx∫ex dxdx+12×sin 2×2-2x⇒y=x ex-ex+14sin 2x-2x+CHence, y=x ex-ex+14sin 2x-2x+C is the solution to the given differential equation.
Q21.
Answer :
We have, x3+x2+x+1dydx=2×2+x⇒ dydx=2×2+xx3+x2+x+1⇒ dy=2×2+xx+1×2+1dxIntegrating both sides, we get∫dy=∫2×2+xx+1×2+1dx⇒y=∫2×2+xx+1×2+1dxLet 2×2+xx+1×2+1=Ax+1+Bx+Cx2+1⇒2×2+x=Ax2+A+Bx2+Bx+Cx+C⇒2×2+x=A+Bx2+B+Cx+A+CComparing the coefficients on both sides, we getA+B=2 …..1B+C=1 …..2A+C=0 …..3Solving 1, 2 and 3, we getA=12B=32C=-12∴y=12∫1x+1dx+∫32x-12×2+1 dx=12∫1x+1dx+12∫3xx2+1dx-12∫1×2+1dx=12∫1x+1dx+34∫2xx2+1dx-12∫1×2+1dx=12logx+1+34logx2+1-12tan-1 x+CHence, y=12logx+1+34logx2+1-12tan-1 x+C is the solution to the given differential equation.
Q22.
Answer :
We have, sin dydx=k⇒dydx=sin-1 k⇒dy=sin-1 kdxIntegrating both sides, we get∫dy=∫sin-1 k dx⇒y=xsin-1 k +C …..1It is given that y0=1.∴1=0×sin-1 k+C⇒C=1Substituting the value of C in 1, we gety=x sin-1 k +1⇒y-1=x sin-1 k Hence, y-1=x sin-1 k is the solution to the given differential equation.
Q23.
Answer :
We have, edydx=x+1Taking log on both sides, we getdydx log e=logx+1⇒dydx=logx+1⇒dy=logx+1dxIntegrating both sides, we get∫dy=∫logx+1dx⇒y=∫1II×log x+1I dx ⇒y=log x+1∫1 dx-∫ddxlog x+1∫1 dxdx⇒y=x log x+1-∫xx+1dx⇒y=x log x+1-∫1-1x+1dx⇒y=x log x+1-x+log x+1+C …..1It is given that y0=3.∴3=0×log 0+1-0+log 0+1+C⇒C=3Substituting the value of C in 1, we gety=x log x+1+log x+1-x+3⇒y=x+1 log x+1-x+3Hence, y=x+1 log x+1-x+3 is the solution to the given differential equation.
Q24.
Answer :
We have, C’ x=2+0.15x⇒dCdx=2+0.15x⇒dC=2+0.15xdxIntegrating both sides, we get∫dC=∫2+0.15x dx⇒C=2x+0.152×2+D …..1It is given that C0=100.∴100=20 +0.1520+D⇒D=100Substituting the value of D in 1, we getC=2x +0.152×2+100Hence, C=2x +0.152×2+100 is the solution to the given differential equation.
Q25.
Answer :
We have, xdydx+1=0⇒ dydx=-1x⇒dy=-1xdxIntegrating both sides, we get⇒∫dy=∫-1xdx⇒y=-logx+C …..1It is given that y-1=0.∴0=-log-1+C⇒C=0Substituting the value of C in 1, we gety=-logxHence, y=-logx is the solution to the given differential equation.
Q26.
Answer :
We have, xx2-1dydx=1⇒ dydx=1xx2-1⇒dy=1xx2-1dxIntegrating both sides, we get∫dy=∫1xx2-1dx⇒y=∫1xx2-1dx⇒y=∫1xx-1x+1dxLet 1xx-1x+1 =Ax+Bx-1+Cx+1⇒1=Ax2-1+Bx2+x+Cx2-x⇒1=A+B+Cx2+B-Cx-AEquating the coefficients on both sides we getA+B+C=0 …..1B-C=0 …..2A=-1 …..3Solving 1, 2 and 3, we getA=-1B=12C=12∴y=12∫1x-1dx-∫1xdx+12∫1x+1dx=12logx-1-logx+12logx+1+C=12logx-1+12logx+1-logx+CIt is given that y2=0.∴0=12log2-1+12log2+1-log2+C⇒C=log2-12log3Substituting the value of C, we gety=12logx-1+12logx+1-logx+log2-12log3⇒2y=logx-1+logx+1-2logx+2log2-log3⇒2y=logx-1+logx+1-logx2+log 4-log 3⇒2y=log4x-1x+13×2⇒y=12log4x2-13x2Hence, y=12log4x2-13×2 is the solution to the given differential equation.
Page 22.37 Ex. 22.6
Q1.
Answer :
We have,dydx+1+y2y=0
⇒dydx=-1+y2y⇒dxdy=-y1+y2⇒dx=-y1+y2dyIntegrating both sides, we get∫dx=∫-y1+y2dy⇒x=∫-y1+y2dyPutting 1+y2=t, we get2y dy=dt∴x=-12∫1tdt⇒x=-12logt+C⇒x=-12log1+y2+C⇒x+12log1+y2=CHence, x+12log1+y2=C is the required solution.
Q2.
Answer :
We have,dydx=1+y2y3
⇒dxdy=y31+y2⇒dx=y31+y2dyIntegrating both sides, we get∫dx=∫y31+y2dy⇒x=∫y+y3-y1+y2dy⇒x=∫1+y2y-y1+y2dy⇒x=∫y dy-∫y1+y2dy⇒x=y22-∫y1+y2dyPutting 1+y2=t we get 2y dy=dt∴x=y22-12∫1tdt⇒x=y22-12logt+C⇒x=y22-12log1+y2+C ∵ t=1+y2Hence, x=y22-12log1+y2+C is the required solution.
Q3.
Answer :
We have,dydx=sin2 y⇒dxdy=1sin2y⇒dx=cosec2y dyIntegrating both sides, we get∫dx=∫cosec2y dy⇒x=-cot y+C⇒x+ cot y=CHence, x+ cot y=C is the required solution.
Q4.
Answer :
We have,dydx=1-cos 2y1+cos 2y⇒dxdy=1+ cos 2y1-cos 2y⇒dx=1+ cos 2y1-cos 2ydy⇒dx=2 cos2y2 sin2ydy⇒dx=cot2y dyIntegrating both sides, we get⇒∫dx=∫cot2y dy⇒x=∫cosec2y-1 dy⇒x=∫cosec2y dy-∫dy⇒x=-cot y-y+C⇒x+ cot y+y=CHence, x+ cot y+y=C is the required solution.
Page 22.53 Ex. 22.7
Q1.
Answer :
We have,x-1dydx=2 xy⇒x-1dy=2xy dx⇒2xx-1dx=1ydyIntegrating both sides, we get2∫xx-1dx=∫1ydy⇒2∫x-1+1x-1dx=∫1ydy⇒2∫dx+2∫1x-1dx=∫1ydy⇒2x+2 logx-1=log y+C
Q2.
Answer :
We have,1+x2 dy=xy dx⇒1ydy=x1+x2dxIntegrating both sides, we get∫1ydy=∫x1+x2dxSubstituting 1+x2=t, we get2x dx=dt∴∫1ydy=12∫1tdt⇒logy=12log t +log C ⇒logy=12log1+x2+log C (∵ t=1+x2)⇒logy=logC1+x2⇒y=C1+x2Hence, y=C1+x2 is the required solution.
Q3.
Answer :
We have,dydx=ex+1 y⇒1ydy=ex+1 dxIntegrating both sides, we get ∫1ydy=∫ex+1 dx⇒log y=ex+x+CHence, log y=ex+x+C is the required solution.
Q4.
Answer :
We have,x-1dydx=2x3y⇒1ydy=2x3x-1dxIntegrating both sides, we get∫1ydy=∫2x3x-1dx⇒log y=2∫x3-1+1x-1dx⇒log y=2∫x-1×2+x+1+1x-1dx⇒log y=2∫x2+x+1dx+2∫1x-1dx⇒log y=23×3+x2+2x+log x-1+CHence, log y=23×3+x2+2x+log x-1+C is the required solution.
Q5.
Answer :
We have,xyy+1dy=x2+1dx⇒yy+1dy=x2+1xdx⇒y2+ydy=x+1xdxIntegrating both sides, we get ∫y2+ydy=∫x+1xdx⇒∫y2 dy+∫ y dy=∫x dx+∫1xdx⇒y33+y22=x22+log x+CHence, y33+y22=x22+log x+C is the required solution.
Q6.
Answer :
We have,5dydx=ex y4⇒5y4dy=exdxIntegrating both sides, we get ∫5y4dy=∫exdx⇒-53y3=ex+CHence, -53y3=ex+C is the required solution.
Q7.
Answer :
We have,x cos y dy=x exlog x+ex dx⇒ cos y dy= ex log x+1xexdxIntegrating both sides, we get∫ cos y dy=∫ ex log x+1xexdx⇒sin y=log x ∫ ex dx-∫1xex dx+∫1xex dx⇒sin y=ex log x+CHence, sin y=ex log x+C is the required solution.
Q8.
Answer :
We have,dydx=ex+y+x2 ey⇒dydx=exey+x2ey⇒dydx=eyex+x2⇒1eydy=ex+x2 dxIntegrating both sides, we get ∫1eydy=∫ex+x2 dx⇒-e-y=ex+x33+C
Q9.
Answer :
We have,xdydx+y=y2⇒xdydx=y2-y⇒1y2-ydy=1xdxIntegrating both sides, we get ∫1y2-ydy=∫1xdx⇒∫1yy-1dy=∫1xdx …..1Let 1yy-1=Ay+By-1⇒1=Ay-1+ByPutting y=0, we get1=-A⇒A=-1Putting y=1, we get1=B∴1yy-1=-1y+1y-1⇒∫1yy-1dy=∫-1y dy+∫1y-1dy …..2 From (1) & (2), we get ∫-1y dy+∫1y-1dy =∫1xdx ⇒-log y+log y-1 =log x+log C⇒log y-1y- log x=log C⇒logy-1xy=log C⇒y-1xy=C⇒y-1=CxyHence, y-1=Cxy is the required solution.
Q10.
Answer :
We have,ey+1 cos x dx +ey sin x dy=0⇒ey sin x dy=-ey+1 cos x dx⇒eyey+1dy=-cos xsin xdx⇒eyey+1dy=-cot x dxIntegrating both sides, we get ∫eyey+1dy=-∫cot x dxPutting ey+1=t, we getey dy=dt∴∫dtt=-∫cot x dx⇒log t =-log sin x +log C ⇒log ey+1+log sin x=log C⇒logey+1 sin x=log C⇒ey+1 sin x=C⇒ey+1 sin x=CHence, ey+1 sin x=C is the required solution.
Q11.
Answer :
We have,x cos 2 y dx=y cos 2x dy⇒xcos2xdx=ycos2ydy⇒x sec2x dx=y sec2y dyIntegrating both sides, we get ∫xI sec2xII dx=∫yI sec2yII dy⇒x∫sec2x dx-∫ddxx∫sec2x dxdx=y∫sec2y dy-∫ddyy∫sec2y dydy⇒x tan x-∫tan x dx=y tan y-∫tan y dy⇒x tan x-log sec x=y tan y-log sec y+C⇒x tan x- y tan y=log sec x-log sec y+CHence, x tan x- y tan y=log sec x-log sec y+C is the required solution.
Q12.
Answer :
We have,xy dy=y-1x+1 dx⇒yy-1dy=x+1xdxIntegrating both sides, we get∫yy-1dy=∫x+1xdx⇒∫y-1+1y-1dy=∫x+1xdx⇒∫dy+∫1y-1dy=∫dx+∫1xdx⇒y+log y-1=x+log x +C⇒y-x=log x- log y-1+CHence, y-x=log x – log y-1+C is the required solution.
Q13.
Answer :
We have,xdydx+cot y=0⇒xdydx=- cot y⇒1xdx=-1cot ydy⇒1xdx=-tan y dyIntegrating both sides, we get∫1xdx=-∫tan y dy⇒ln x =-ln sec y+ln C⇒ln x =ln cos y +ln C⇒x=C cos y Hence, x=C cos y is the required solution.
Q14.
Answer :
We have,dydx=xex logx +exx cos y⇒x cos y dy=x exlog x+ex dx⇒ cos y dy= ex log x+1xexdxIntegrating both sides, we get∫ cos y dy=∫ ex log x+1xexdx⇒sin y=log x ∫ ex dx-∫1xex dx+∫1xex dx⇒sin y=ex log x+CHence, sin y=ex log x+C is the required solution.
Q15.
Answer :
We have,dydx=ex+y+ey x3⇒dydx=exey+eyx3⇒dydx=eyex+x3⇒ex+x3 dx=1eydyIntegrating both sides, we get ∫ex+x3 dx=∫1eydy⇒ex+x44=-e-y+C⇒ex+e-y+x44=CHence,ex+e-y+x44=C is the required solution.
Q16.
Answer :
We have,y1+x2+x1+y2dydx=0⇒x1+y2dydx=-y1+x2⇒x1+y2dy=-y1+x2 dx⇒1+y2ydy=-1+x2xdx
Integrating both sides, we get∫1+y2ydy=-∫ 1+x2xdxPutting 1+y2=t2 and 1+x2=u2, we get2y dy=2t dt and 2x dx=2u du⇒dy=tydt and dx=uxdu∴∫t2y2dt=-∫u2x2dx⇒∫t2t2-1dt=-∫u2u2-1du
⇒∫t2-1+1t2-1dt=-∫u2-1+1u2-1du⇒∫dt+∫1t2-1dt=-∫du-∫1u2-1du⇒t+12logt-1t+1=-u-12logu-1u+1+CSubstituting t by 1+y2 and u by 1+x2
1+y2+12log1+y2-11+y2+1=-1+x2-12log1+x2-11+x2+1+C⇒1+y2+1+x2+12log1+x2-11+x2+1+12log1+y2-11+y2+1=CHence, 1+y2+1+x2+12log1+x2-11+x2+1+12log1+y2-11+y2+1=C is the required solution.
Q17.
Answer :
We have,1+x2 dy+1+y2 dx=01+x2 dy=-1+y2 dx11+y2 dy=-11+x2 dxIntegrating both sides, we get∫11+y2 dy=-∫11+x2 dx⇒log y+1+y2=-log x+1+x2+log C⇒log y+1+y2+log x+1+x2=log C⇒log y+1+y2x+1+x2=log C⇒y+1+y2x+1+x2=CHence, log y+1+y2x+1+x2=C is the required differential equation.
Q18.
Answer :
We have,1+x2+y2+x2y2+xydydx=0⇒1+x21+y2+xydydx=0⇒xydydx=-1+x21+y2⇒xydydx=-1+x21+y2⇒y1+y2dy=-1+x2xdxIntegrating both sides, we get⇒∫y1+y2dy=-∫1+x2xdx⇒∫y1+y2dy=-∫x1+x2x2dxPutting 1+y2=t and 1+x2=u2⇒2y dy=dt and 2x dx=2udu⇒y dy=dt2 and xdx=udu∴Integral becomes,12∫dtt=-∫u×uu2-1du⇒t=-∫u2u2-1du⇒t=-∫u2u2-1du⇒t=-∫1+1u2-1du⇒t=-∫(1)du-∫1u2-1du⇒t=-u-12logu-1u+1+C⇒1+y2=-1+x2-12log1+x2-11+x2+1+C⇒1+y2+1+x2+12log1+x2-11+x2+1=C
Q19.
Answer :
dydx=exsin2x+sin 2xy2log y+1⇒y2log y+1dy=exsin2x+sin 2xdx⇒2y log y+ydy=exsin2x+exsin 2xdx⇒2y log y dy + y dy=exsin2x dx+exsin 2x dxIntegrating both sides, we get2∫yII log yI dy+∫y dy=∫ex IIsin2xI dx+∫exsin 2x dx⇒2log y∫y dy-∫ddylog y∫y dydy+∫y dy=sin2x∫ex dx-∫ddxsin2x∫ex dxdx+∫exsin 2x dx⇒2log y y22-∫1yy22dy+∫y dy=sin2x ex-∫2sin xcos x exdx+∫exsin 2x dx+C⇒y2log y-∫y dy+∫y dy=exsin2x-∫exsin 2x dx+∫exsin 2x dx+C⇒y2log y=exsin2x+C
Q20.
Answer :
We have,dydx=x2 log x+1sin y+y cos y
⇒sin y+y cos y dy=x2 log x +1 dxIntegrating both sides, we get∫sin y+y cos y dy=∫x2 log x +1 dx⇒∫sin y dy+∫y cos y dy=2∫x log x dx+∫x dx⇒-cos y+y∫cos y dy-∫ddyy∫ cos y dydy=2log x∫ x dx-∫ddxlog x∫x dxdx+x22⇒-cos y+y sin y-∫sin y dy=2 log x× x22-∫1x×x22 +x22⇒ -cos y+y sin y+cos y=x2 log x-x22+x22+C⇒ y sin y=x2 log x+CHence, y sin y=x2 log x+C is the required solution.
Q21.
Answer :
We have,1-x2 dy+xy dx= xy2 dx ⇒1-x2 dy=xy2 dx- xy dx⇒1-x2 dy=xy y-1 dx⇒1yy-1 dy=x1-x2 dxIntegrating both sides, we get∫1yy-1 dy=∫x1-x2 dx …..(1)Considering LHS of (1),Let 1yy-1=Ay+By-1⇒1=Ay-1+By …..(2) Substituting y=1 in (2),1=B Substituting y=0 in (2),1=-A⇒A=-1Substituting the values of A and B in 1yy-1=Ay+By-1, we get1yy-1=-1y+1y-1⇒∫1yy-1dy=∫-1ydy+∫1y-1dy =-log y+ log y-1+C1 Now, considering RHS of (2), we have∫x1-x2 dxHere, putting 1-x2=t, we get -2x dx=dt∴∫x1-x2 dx=-12∫1tdt =-12log t+C2 =-12log 1-x2+C2 ∵ t=1-x2Now, substituting the value of ∫1yy-1dy and ∫x1-x2 dx in (1), we get-log y+log y-1+C1=-12log 1-x2+C2⇒-log y+log y-1=-12log 1-x2+C whereC=C2-C1
Q22.
Answer :
We have,tan y dx+sec2y tanx dy=0⇒sec2y tanx dy=-tan y dx⇒sec2ytan y dy=-1tanxdx⇒1cos2y×cosysinydy=-cot x dx⇒1siny cosydy=-cot x dx⇒2sin 2y dy=-cot x dx⇒2 cosec 2y dy=-cot x dxIntegrating both sides, we get2∫cosec 2y dy=-∫cot x dx⇒log tan x=-log sin x= log C⇒log tan x+log sin x= log C⇒log tan x×sin x= log C⇒tan x×sin x=C
Q23.
Answer :
We have,1+x1+y2 dx+1+y1+x2dy=0⇒1+x1+y2 dx=-1+y1+x2dy⇒1+x1+x2dx=-1+y1+y2dyIntegarting both sides, we get ∫1+x1+x2dx=-∫1+y1+y2dy⇒∫11+x2dx+∫x1+x2dx=-∫11+y2dy-∫y1+y2dySubstituting 1+x2=t in the second integral of LHS and 1+y2=u in the second integral of RHS, we get2x dx=dt and 2ydy=du∴∫11+x2dx+12∫1tdt=-∫11+y2dy-12∫1udu⇒ tan-1x+12log t=-tan-1 y-12log u+C⇒tan-1x+12log 1+x2=-tan-1 y-12log 1+y2+C⇒tan-1x+tan-1y+12log 1+x2+12log 1+y2=C⇒tan-1x+tan-1 y+12log 1+x21+y2=CHence, tan-1x+tan-1 y+12log 1+x21+y2=C is the required solution.
Q24.
Answer :
We have,tan y dydx=sin x+y+sin x-y⇒tan y dydx=sin x cos y+cos x sin y+sin x cos y-cos x sin y⇒tan y dydx=2 sin xcos y ⇒tan ycos ydy=2 sin x dx⇒tany sec y dy=2 sinx dxIntegrating both sides, we get ∫tany sec y dy=2∫sin x dx⇒sec y=-2 cos x+C⇒sec y+2 cos x=CHence, sec y+2 cos x=C is the required solution.
Q25.
Answer :
We have,cos x cos y dydx=-sin x siny ⇒cos ysin ydy=-sin xcos xdx⇒cot y dy=-tan x dxIntegrating both sides, we get ∫cot y dy=-∫tan x dx⇒log sin y=-log sec x +log C⇒log sin y=log cos x +log C⇒sin y=C cos xHence, sin y=C cos x is the required solution.
Q26.
Answer :
We have,dydx+cos x sin ycos y=0⇒dydx=-cos x sin ycos y⇒cos ysin ydy=-cos x dx⇒cot y dy=-cos x dxIntegrating both sides, we get ∫cot y dy=-∫cos x dx⇒log sin y=-sin x+CHence, log sin y=-sin x+C is the required solution.
Q27.
Answer :
We have,x1-y2 dx+y1-x2 dy=0⇒y1-x2 dy=-x1-y2 dx⇒y1-y2 dy=-x1-x2 dxIntegrating both sides, we get∫y1-y2 dy=-∫x1-x2 dxSubstituting 1-y2=t and 1-x2=u, we get-2y dy=dt and -2x dy=du∴ -12∫1tdt=12∫1udu⇒-t12=u12+K⇒1-x2+1-y2=-K⇒1-x2+1-y2=C where, C=-KHence, 1-x2+1-y2=C is the required solution.
Q28.
Answer :
We have,y1+ex dy=y+1 ex dx⇒yy+1dy=ex1+exdxIntegrating both sides, we get ∫yy+1dy=∫ex1+exdxSubstituting 1+ex=t, we getexdx=dt∴∫yy+1dy=∫1tdt⇒∫y+1-1y+1dy=∫1tdt⇒∫dy-∫1y+1dy=∫1tdt⇒y-log y+1=log t+C⇒y-log y+1=log 1+ex+C
Q29.
Answer :
We have,y+xy dx+x-xy2 dy=0⇒y1+xdx=xy2-1dy⇒1+xxdx=y2-1ydyIntegrating both sides, we get ∫1+xxdx=∫y2-1ydy⇒∫1xdx+∫dx=∫y dy-∫1ydy⇒log x+x=y22-log y+C⇒log x+x-y22+log y=CHence, log x+x-y22+log y=C is the required solution.
Q30.
Answer :
We have,dydx=1-x+y-xy⇒dydx=1+y-x1+y⇒dydx=1+y1-x⇒11+ydy=1-x dxIntegrating both sides, we get∫11+ydy=∫1-x dx⇒log 1+y=x-x22+CHence, log 1+y=x-x22+C is the required solution.
Page 22.54 Ex. 22.7
Q31.
Answer :
We have,y2+1 dx-x2+1 dy=0⇒y2+1 dx=x2+1 dy⇒1×2+1dx=1y2+1dyIntegrating both sides, we get∫1×2+1dx=∫1y2+1dy⇒tan-1 x=tan-1 y+C⇒tan-1 x-tan-1y=CHence, tan-1 x-tan-1y=C is the required solution.
Q32.
Answer :
We have,dy+x+1y+1 dx=0⇒dy=-x+1y+1 dx⇒1y+1dy=-x+1 dxIntegrating both sides, we get∫1y+1dy=-∫x+1 dx⇒log y+1=-x22-x+C⇒log y+1+x22+x=CHence, log y+1+x22+x=C is the required solution.
Q33.
Answer :
We have,dydx=1+x21+y2⇒11+y2dy=1+x2 dx Integrating both sides, we get∫11+y2dy=∫1+x2 dx ⇒tan-1y=x+x33+CHence, tan -1y=x+x33+C is the required solution.
Q34.
Answer :
We have,x-1dydx=2×3 y⇒1ydy=2x3x-1dxIntegrating both sides, we get ∫1ydy=∫2x3x-1dx⇒log y=2∫x3-1+1x-1dx⇒log y=2∫x3-1x-1dx+∫1x-1dx⇒log y=2∫x-1×2+x+1x-1dx+∫1x-1dx⇒log y=2∫x2+x+1 dx+∫1x-1dx⇒log y=2 x33+x22+x+log x-1+C⇒log y=23×3+x2+2x+log x-12+C⇒y=e23x3+x2+2x+log x-12+C⇒y=eC×elog x-12×e23x3+x2+2x⇒y=C1x-12 e23x3+x2+2x ∵ eln x=x and where, C1=eC ∴ y=C1x-12 e23x3+x2+2x is required solution.
Q35.
Answer :
We have,dydx=ex+y+e-x+y⇒dydx=eyex+e-x⇒e-ydy=ex+e-x dxIntegrating both sides, we get∫e-ydy=∫ex+e-x dx⇒-e-y=ex-e-x+C⇒e-x-e-y=ex+CHence, e-x-e-y=ex+C is the required solution.
Q36.
Answer :
We have,dydx=cos2 x-sin2 x cos2 y⇒dydx=cos 2x cos2y⇒1cos2ydy=cos 2x dx⇒sec2y dy=cos 2x dxIntegrating both sides, we get∫sec2y dy=∫cos 2x dx⇒tan y=sin 2×2+CHence, tan y=sin 2×2+C is the required solution.
Q37.
Answer :
We have,xy2+2x dx+x2y+2y dy=0⇒xy2+2 dx+yx2+2 dy=0⇒xy2+2 dx=-yx2+2 dy⇒xx2+2 dx=-yy2+2 dyIntegrating both sides, we get∫xx2+2 dx=-∫yy2+2 dy⇒12∫2xx2+2 dx=-12∫2yy2+2 dy⇒12log x2+2=-12log y2+2+log C⇒12log x2+2+12log y2+2=log C⇒log x2+2+log y2+2=2log C⇒log x2+2y2+2=log C2⇒x2+2y2+2=C2⇒x2+2y2+2=K⇒y2+2=Kx2+2
Q38.
Answer :
(i) We have, xydydx=1+x+y+xy⇒xydydx=1+x1+y⇒y1+ydy=1+xxdxIntegrating both sides, we get ∫y1+ydy=∫1+xxdx⇒∫1+y-11+ydy=∫1+xxdx⇒∫dy-∫11+ydy=∫1xdx+∫dx⇒y-log 1+y= log x+x+C⇒y= log x+log 1+y+x+C⇒y=log x1+y+x+C Hence, y=log x1+y+x+C is the required solution.
(ii) We have,y1-x2dydx=x1+y2⇒y1+y2dy=x1-x2dxIntegrating both sides ,∫y1+y2dy=∫x1-x2dxSubstituting 1+y2=t and 1-x2=u 2ydy=dt and -2x dx=du∴12∫1tdt=-12∫1udu⇒12log t =-12log u+log C⇒12log 1+y2=-12log 1-x2+log C⇒12log 1+y2+log 1-x2=log C⇒log 1+y21-x2=2 log C⇒ 1+y21-x2=C2 ⇒1+y21-x2=C1 , where C1=C2Hence,1+y21-x2=C1 is the required solution.
Q39.
Answer :
We have,dydx=y tan 2x, y0=2⇒1ydy=tan 2x dxIntegrating both sides, we get ∫1ydy=∫tan 2x dx⇒log y=12log sec 2x+12log C⇒y2=C sec 2x …..1It is given that at x=0, y=2.∴ C=4Substituting the value of C in (1), we get∴ y2 =4cos 2x⇒y=2cos 2x Hence, y=2cos 2x is the required solution.
Q40.
Answer :
We have,2xdydx=3y, y1=2⇒2ydy=3xdxIntegrating both sides, we get 2∫1ydy=3∫1xdx⇒2 log y=3 log x+log C⇒log y2=log x3+log C⇒y2=Cx3 ….(1)It is given that at x=1, y=2.Substituting the values of x and y in (1), we getC=4Now, substituting the value of C in (1), we get⇒y2=4x3Hence, y2=4×3 is the required solution.
Q41.
Answer :
We have,xydydx=y+2, y2=0⇒yy+2dy=1xdxIntegrating both sides, we get∫yy+2dy=∫1xdx⇒∫y+2-2y+2dy=∫1xdx⇒∫dy-2∫1y+2dy=log x+C⇒y-2 log y+2=log x+C …..(1) It is given that at x=2, y=0.Substituting the values of x and y in (1), we get-2log 2-log 2=C⇒-log 22×2=C⇒C=-log 8Substituting the value of C in (1), we gety-2 log y+2=log x-log 8⇒y-2 log y+2=log x8Hence, y-2log y+2=log x8 is the required solution.
Q42.
Answer :
We have,dydx=2ex y3, y0=12⇒1y3dy=2exdxIntegrating both sides, we get ∫1y3dy=∫2exdx⇒-12y2=2ex+C …..(1)Given: at x=0, y=12Substituting the values of x and y in (1), we get-12×14=2e0+C⇒C=-2-2⇒C=-4Substituting the value of C in (1), we get⇒-12y2=2ex-4⇒y28-4ex=1Hence, y28-4ex=1 is the required solution.
Q43.
Answer :
We have,drdt=-rt, r0=r0
⇒1rdr =-t dtIntegrating both sides, we get∫1rdr =-∫t dt⇒log r=-t22+C ….(1)Given: t=0, r=r0. Substituting the values of x and y in (1), we getlog r0 =0+C⇒C=log r0 Substituting the value of C in (1), we get
log r=-t22+log r0 ⇒log r -log r0 =-t22⇒log rr0=-t22⇒r=r0e-t22Hence, r=r0e-t22 is the required solution.
Q44.
Answer :
We have,dydx=y sin2x, y0=1⇒1ydy =sin 2x dxIntegrating both sides, we get∫1ydy =∫sin 2x dx⇒log y=-cos 2×2+C …..(1)Given: x=0, y=1.Substituting the values of x and y in (1), we getlog 1 =-12+C⇒C=12Substituting the value of C in (1), we get
log y=-cos 2×2+12⇒log y =1-cos 2×2⇒log y=sin 2x⇒y=esin2xHence, y=esin2x is the required solution.
Q45.
Answer :
(i) dydx=y tanx, y0=1⇒1ydy =tan x dxIntegrating both sides, we get∫1ydy =∫tan x dx⇒log y=log sec x+C …..(1)We know that at x=0 and y=1.Substituting the values of x and y in (1), we getlog 1 =log 1 +C⇒C=0Substituting the value of C in (1), we get
log y=log sec x+0⇒y= sec xHence, y=sec x, where x∈-π2,π2, is the required solution.
Q46.
Answer :
We have,xdydx+cot y=0⇒xdydx=-cot y⇒tan y dy=-1xdxIntegrating both sides, we get∫tan y dy=-∫1xdx⇒log sec y=- log x+log C⇒log x sec y =log C⇒x sec y=C …..(1) Given: x=2 , y= π4.Substituting the values of x and y in (1), we get2 sec π4=C⇒C=2Substituting the value of C in (1), we getx sec y=2⇒x=2 cos yHence, x=2 cos y is the required solution.
Q47.
Answer :
We have,1+x2dydx+1+y2=0 , y=1 when x=0⇒1+x2dydx=-1+y2⇒11+y2 dy=-11+x2dxIntegrating both sides, we get∫11+y2 dy=-∫11+x2dx⇒tan-1y=-tan-1x+C⇒tan-1y+tan-1x=C …..(1) Given: x=0, y= 1.Substituting the values of x and y in (1), we get π4+0=C⇒C=π4Substituting the value of C in (1), we gettan-1y+tan-1x=π4⇒tan-1x+tan-1y=π4⇒tan-1x+y1-xy=π4⇒x+y1-xy=1⇒x+y=1-xyHence, x+y=1-xy is the required solution.
Q48.
Answer :
We have,dydx=2xlog x+1sin y+ycos y⇒sin y+ycos y dy=2xlog x+1 dxIntegrating both sides, we get∫sin y+ycos y dy=∫2xlog x+1 dx⇒∫sin y dy+∫ ycos y dy=∫2x log x dx+∫2x dx⇒-cos y+y∫cos y dy-∫ddyy∫cos y dydy=2log x ∫x dx-∫ddxlog x∫x dxdx+x2⇒-cos y+y sin y+cos y=2 log x×x22-x24+x2+C⇒y sin y=x2log x-x22+x2+C⇒y sin y=x2log x+x22+C …..(1)Given: x=1, y=0.Substituting the values of x and y in (1), we get 0=0+12+C⇒C=-12Substituting the value of C in (1), we gety sin y=x2log x+x22-12⇒2y sin y=2x2log x+x2-1Hence, 2y sin y=2x2log x+x2-1 is the required solution.
Q49.
Answer :
We have,edydx=x+1⇒dydx=log x+1⇒dy=log x+1 dxIntegrating both sides, we get ∫dy=∫log x+1 dx⇒y=log x+1∫1 dx-∫ddxlog x+1∫1 dxdx⇒y= x log x+1-∫1x+1×x dx⇒y= x log x+1-∫1-1x+1 dx⇒y= x log x+1-∫dx+∫1x+1dx⇒y= x log x+1-x+log x+1+C⇒y= x+1 log x+1-x+C …..(1)It is given that at x=0 and y=3.Substituing the values of x and y in (1), we get C=3Therefore, substituting the value of C in (1), we gety=x+1 log x+1-x+3Hence, y=x+1 log x+1-x+3 is the required solution.
Q50.
Answer :
We have,cos y dy+ cos x sin y dx=0⇒cos y dy=- cos x sin y dx⇒cot y dy=-cos x dxIntegrating both sides, we get∫cot y dy=-∫cos x dx⇒log sin y=- sin x+C⇒log sin y+ sin x=C ….(1)It is given that at x=π2, y=π2.Substitutuing the values of x and y in 1, we getlog sin π2+ sin π2=C⇒C=1Therefore, substituting the value of C in (1), we get log siny+ sin x=1Hence, log sin y+sin x=1 is the required solution.
Q51.
Answer :
We have,dydx=-4xy2⇒1y2dy=-4x dxIntegrating both sides, we get∫1y2dy=-4∫x dx ⇒-1y=-4×x22+C⇒-1y=-2×2+C …..(1)It is given that at x=0, y=1.Substituting the values of x and y in (1), we getC=-1Therefore, substituting the value of C in (1), we get -1y=-2×2-1⇒y=12×2+1Hence, y=12×2+1 is the required solution.
Q52.
Answer :
We have to find the equation of the curve that passes through the point (0,0)and whose differential equation is dydx=ex sin x.dy=ex sin x dxIntegarting both sides, we get ∫dy=∫ex sin x dx⇒y=∫ex sin x dx …..1⇒y=ex∫ sin x dx-∫ddxex ∫sin x dx dx⇒y=-ex cos x+∫excos x dx⇒y=-ex cos x+ex ∫ cos x dx-∫ddxex ∫cos x dxdx⇒y=-ex cos x+ex sin x-∫exsin x dx⇒y=-ex cos x+ex sin x-y+C Using 1⇒2y=ex sin x-cos x+C …..2The curve passes through the point (0,0)When, x=0; y=0Substituting the value of x and y in 2, we get0=1 0-1+C⇒C=1Substituting the value of C in 2, we get2y=ex sin x-cos x+1Required equation of curve is 2y=ex sin x-cos x+1
Q53.
Answer :
We have,xydydx=x+2y+2⇒yy+2dy=x+2xdxIntegrating both sides, we get∫yy+2dy=∫x+2xdx⇒∫dy-2∫1y+2dy=∫dx+2∫1xdx⇒y-2 log y+2=x+2 log x+C …..(1)This equation represents the family of solution curves of the given differential equation. We have to find a particular member of the family, which passes through the point 1,-1.Substituting x=1 and y=-1 in (1), we get-1-2 log 1=1+2 log 1+C⇒C=-2Putting C=-2 in (1), we get y-2 log y+2=x+2 log x-2 ⇒y-x+2=log x2 y+22 Hence, y-x+2=log x2 y+22 is the equation of the required curve.
Q54.
Answer :
Let r be the radius and V be the volume of the balloon at any time ‘t’.
Then, we have,
V=43πr3Given:dVdt=-k where k>0⇒ddt43πr3=-k⇒4πr2 drdt=-k⇒4πr2dr=-kdt Integrating both sides, we get∫4πr2dr=-∫kdt 43πr3=-kt+C …(1)It is given that at t=0, r=3. Substituting t=0 and r=3 in (1), we get C=36πPutting C=36π in (1), we get 43πr3=-kt+36π …(2)It is also given that at t=3, r=6. Putting t=3 and r=6 in (1), we get288 π=-3k+36π⇒k=-84πPutting k=-84 π in (2), we get43πr3=84π t+36 π⇒r3=63 t+27⇒r=63 t+2713
Page 22.55 Ex. 22.7
Q55.
Answer :
Let P be the principal at any instant t.
Given:
dPdt=r100P⇒dPP=r100dtIntegrating both sides, we get∫dPP=∫r100dt⇒log P=rt100+C ……(1)Initially, i.e. at t=0, let P= P0. Putting P=P0, we get log P0=C, Putting C=log P0 in (1), we getlog P=rt100+log P0⇒log PP0=rt100Substituting P0=100, P=2P0=200 and t=10 in (2), we get log 2 =r10∴r=10 log 2 =10× 0.6931 =6.931
Q56.
Answer :
Let at any instant t, the principal be P.Here, it is given that the principal increases at the rate of 5% per year.dPdt=5P100⇒dPP=120dtIntegrating both sides, we get ln P=t20+ln C ….(1) Initially at t=0, it is given that P=Rs 1000.ln 1000=ln CSubstituting the value of ln C in (1), we get ln P=t20+ln 1000Putting t=10, we getln P1000=0.5⇒P1000=e0.5⇒P=1000×1.648 =1648Therefore, Rs 1000 will be worth Rs 1648 after 10 years.
Q57.
Answer :
Let at any time the bacteria count be N.Given: dNdtα N⇒dNdt=λN⇒1NdN=λdtIntegrating both sides, we get∫1NdN=∫λdt⇒ln N=λt+ln C …(1)Given:at t=0, N=100000therefore, ln C=ln 100000Putting the value in (1) we get,ln N=λt+ln 100000Also, at t=2N=110000Putting the values of t and N in (1), we getln 110000=2λ+ ln 100000⇒12ln 1110=λSubstituting the values of ln C and λ in (1), we getln N=12ln 1110t+ ln 100000 ….(2)When N=200000, let t=T.Substituting these values in (2), we get ln 200000=T2ln 1110+ln 100000⇒ln 2=T2ln 1110⇒T=2ln 2ln 1110Therefore, in 2ln 2ln 1110 hours, the count will reach 200000.
Page 22.63 Ex. 22.8
Q1.
Answer :
We have,dydx=x+y+12Putting x+y+1=v⇒1+dydx=dvdx⇒dydx=dvdx-1∴dvdx-1=v2⇒dvdx=v2+1⇒1v2+1dv=dxIntegrating both sides, we get∫1v2+1dv=∫dx⇒tan-1 v=x+C⇒tan-1x+y+1=x+C
Q2.
Answer :
We have,dydxcosx-y=1⇒dydx=1cosx-yPutting x-y=v⇒1-dydx=dvdx⇒dydx=1-dvdx∴1-dvdx=1cos v⇒dvdx=1-1cos v⇒dvdx=cos v-1cos v⇒cos vcos v-1dv=dxIntegrating both sides, we get∫cos vcos v-1dv=∫dx⇒-∫cos v1+cos v1-cos2 vdv=∫dx⇒-∫cos v1+cos vsin2 vdv=∫dx⇒-∫cot v cosec v+cot2vdv=∫dx⇒-∫cot v cosec v+cosec2 v-1dv=∫dx⇒–cosec v-cot v-v=x+C⇒cosec x-y+cot x-y+x-y=x+C⇒cosec x-y+cot x-y-y=C⇒1+cos x-ysin x-y-y=C⇒cotx-y2=y+C
Q3.
Answer :
We have,dydx=x-y+32x-y+5Putting x-y=v⇒1-dydx=dvdx⇒dydx=1-dvdx∴1-dvdx=v+32v+5⇒dvdx=1-v+32v+5⇒dvdx=2v+5-v-32v+5⇒dvdx=v+22v+5⇒2v+5v+2dv=dxIntegrating both sides, we get∫2v+5v+2dv=∫dx⇒∫2v+4+1v+2dv=∫dx⇒∫2v+4v+2+1v+2dv=∫dx⇒2∫dv+∫1v+2dv=∫dx⇒2v+log v+2=x+C⇒2x-y+logx-y+2=x+C
Q4.
Answer :
We have,dydx=x+y2Let x+y=v⇒1+dydx=dvdx⇒dydx=dvdx-1∴dvdx-1=v2⇒dvdx=v2+1⇒1v2+1dv=dxIntegrating both sides, we get∫1v2+1dv=∫dx⇒tan-1 v=x+C⇒v=tanx+C⇒x+y=tanx+C
Q5.
Answer :
We have,x+y2dydx=1⇒dydx=1x+y2Let x+y=v⇒1+dydx=dvdx⇒dydx=dvdx-1∴dvdx-1=1v2⇒dvdx=1v2+1⇒v2v2+1dv=dxIntegrating both sides, we get∫v2v2+1dv=∫dx⇒∫v2+1-1v2+1dv=∫dx⇒∫1-1v2+1dv=∫dx⇒v-tan-1 v=x+C⇒x+y-tan-1 x+y=x+C⇒y-tan-1 x+y=C
Q6.
Answer :
We have,cos2x-2y=1-2dydx⇒2dydx=1-cos2x-2yLet x-2y=v⇒1-2dydx=dvdx⇒2dydx=1-dvdx∴1-dvdx=1-cos2 v⇒dvdx=cos2 v⇒sec2 v dv=dxIntegrating both sides, we get∫sec2 v dv=∫dx⇒tan v=x-C⇒tanx-2y=x-C⇒x=tanx-2y+C
Q7.
Answer :
We have,dydx=secx+ydydx=1cosx+yLet x+y=v⇒1+dydx=dvdx⇒dydx=dvdx-1∴dvdx-1=1cos v⇒dvdx=cos v+1cos v⇒cos vcos v+1dv=dxIntegrating both sides, we get∫cos vcos v+1dv=∫dx⇒∫cos v1-cos v1-cos2 vdv=∫dx⇒∫cos v1-cos vsin2 vdv=∫dx⇒∫cos v-cos2 vsin2 vdv=∫dx⇒∫cot v cosec v-cot2 vdv=∫dx⇒∫cot v cosec v-cosec2 v+1dv=∫dx⇒-cosec v+cot v+v=x+C⇒-cosec x+y+cot x+y+x+y=x+C⇒-cosec x+y+cot x+y+y=C⇒-1+cos x+ysin x+y+y=C⇒-tanx+y2+y=C⇒y=tanx+y2+C
Q8.
Answer :
We have,dydx=tanx+ydydx=sinx+ycosx+yLet x+y=v∴ 1+dydx=dvdx⇒dydx=dvdx-1∴dvdx-1=sin vcos v⇒dvdx=sin vcos v+1⇒dvdx=sin v+cos vcos v⇒cos vsin v+cos vdv=dxIntegrating both sides, we get∫cos vsin v+cos vdv=∫dx⇒12∫sin v+cos v+cos v-sin vsin v+cos vdv=∫dx⇒12∫dv+12∫cos v-sin vsin v+cos vdv=∫dx⇒12v+12∫cos v-sin vsin v+cos vdv=xPutting sin v+cos v=t⇒cos v-sin vdv=dt∴12v+12∫dtt=x⇒12v+12log t=x+C⇒12x+y+12log sin x+y+cos x+y=x+C⇒12y-x+12log sin x+y+cos x+y=C⇒y-x+log sin x+y+cos x+y=2C⇒y-x+log sin x+y+cos x+y=K where, K=2C
Q9.
Answer :
We have,
(x + y) (dx − dy) = dx + dy
⇒x dx +y dx -x dy- y dy= dx+dy⇒x+y-1dx=x+y+1dy⇒dydx=x+y-1x+y+1Let x+y=v∴ 1+dydx=dvdx⇒dydx=dvdx-1∴ dvdx-1=v-1v+1⇒dvdx=v-1v+1+1⇒dvdx=v-1+v+1v+1⇒dvdx=2vv+1⇒v+12vdv=dxIntegrating both sides, we get∫v+12vdv=∫dx⇒12∫dv+12∫1vdv=∫dx⇒12v+12logv=x+C⇒12x+y+12logx+y=x+C⇒12y-x+12logx+y=C
Q10.
Answer :
We have,
x+y+1dydx=1⇒dydx=1x+y+1
Let x+y+1=v∴ 1+dydx=dvdx⇒dydx=dvdx-1∴dvdx-1=1v⇒dvdx=1v+1⇒vv+1dv=dxIntegrating both sides, we get∫vv+1dv=∫dx⇒∫v+1-1v+1dv=∫dx⇒∫1-1v+1dv=∫dx⇒v-logv+1=x+K⇒x+y+1-logx+y+1+1=x+K⇒y-logx+y+2=K-1⇒y-logx+y+2=C1 C1=K-1⇒y-C1=logx+y+2⇒ey-C1=x+y+2⇒eyeC1=x+y+2⇒e-C1ey=x+y+2⇒Cey=x+y+2 C=e-C1⇒x=Cey-y-2
Page 22.80 Ex. 22.9
Q1.
Answer :
We have,x2dy+yx+y dx=0⇒x2dy=-yx+y dx⇒dydx=-yx+yx2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=-vxx+vxx2⇒v+xdvdx=-v1+v⇒xdvdx=-v-v-v2⇒xdvdx=-v2+2v⇒dvv2+2v=-dxx⇒dvvv+2=-dxxIntegrating both sides, we get ∫dvvv+2=-∫dxx⇒12∫1v-1v+2dv=-∫dxx⇒12∫1vdv-∫1v+2dv=-∫dxx⇒12log v-log v+2=-log x +log C⇒12log vv+2=log Cx ⇒log vv+2=2log Cx⇒log vv+2=log C2x2⇒vv+2=C2x2⇒yxyx+2=C2x2⇒yy+2x=C2x2⇒x2y=C2y+2x⇒x2y=Ky+2x Where, K=C2
Q2.
Answer :
We have,dydx=y-xy+xThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=vx-xvx+x⇒v+xdvdx=xv-1xv+1⇒xdvdx=v-1v+1-v⇒xdvdx=v-1-v2-vv+1⇒xdvdx=-v2+1v+1⇒v+1v2+1dv=-1xdxIntegrating both sides, we get∫v+1v2+1dv=-∫1xdx⇒∫vv2+1dv+∫1v2+1dv=-∫1xdx⇒12∫2vv2+1dv+∫1v2+1dv=-∫1xdx⇒12log v2+1+tan-1 v=-log x+C⇒12log v2+1+log x+tan-1 v=C⇒log v2+1+2 log x+2 tan-1 v=2C⇒log v2+1+log x2+2 tan-1 v=2C⇒log v2+1 x2+2 tan-1 v=2C Substituting v=yx, we get⇒log y2x2+1 x2+2 tan-1 yx=2C ⇒log y2+x2 +2 tan-1 yx=k where k=2C
Q3.
Answer :
We have,dydx=y2-x22xyThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=v2x2-x22vx2⇒xdvdx=v2-12v-v⇒xdvdx=-v2+12v⇒2vv2+1dv=-1xdxIntegrating both sides, we get∫2vv2+1dv=-∫1xdxlog v2+1=-log x+log C⇒log v2+1=log Cx⇒v2+1=CxPutting v=yx, we get⇒yx2+1=Cx ⇒ y2+x2=Cx Hence, x2+y2=Cx is the required solution.
Q4.
Answer :
We have,xdydx=x+y⇒dydx=x+yxThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=x+vxx⇒v+xdvdx=1+v⇒xdvdx=1+v-v⇒xdvdx=1⇒dv=1xdxIntegrating both sides, we get ∫dv=∫1xdx⇒v=log x+CPutting v=yx, we get⇒yx=log x+C ⇒y=xlog x+Cx Hence, y=xlog x+Cx is the required solution.
Q5.
Answer :
We have,x2-y2 dx-2xy dy=0⇒dydx=x2-y22xyThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=x2-vx22xvx⇒v+xdvdx=x2-v2x22vx2⇒v+xdvdx=1-v22v⇒xdvdx=1-v22v-v⇒xdvdx=1-3v22v⇒2v1-3v2dv=1xdxIntegrating both sides, we get ∫2v1-3v2dv=∫1xdx⇒-13∫-6v1-3v2dv=∫1xdx⇒-13log 1-3v2=log x+log C⇒log 1-3v2=-3log Cx⇒log 1-3v2=log 1Cx3⇒1-3v2=1Cx3Putting v=yx, we get1-3yx2=1Cx3⇒x2-3y2x2=1C3x3⇒xx2-3y2=1C3⇒xx2-3y2=K where, K=1C3
Q6.
Answer :
We have,dydx=x+y x-yThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=x+vx x-vx⇒v+xdvdx=1+v 1-v⇒xdvdx=1+v1-v-v⇒xdvdx=1+v21-v1-v1+v2dv=1xdxIntegrating both sides, we get∫1-v1+v2dv=∫1xdx⇒∫11+v2dv-∫v1+v2dv=∫1xdx⇒∫11+v2dv-12∫2v1+v2dv=∫1xdxtan-1v-12log 1+v2=log x+CPutting v=yx, we get⇒tan-1yx-12log 1+y2x2=log x+C⇒tan-1yx=12log 1+y2x2+log x+C ⇒tan-1yx=12log x2+y2x2+log x+C ⇒tan-1yx=12log x2+y2-12log x2+log x+C ⇒tan-1yx=12log x2+y2-log x+log x+C ⇒tan-1yx=12log x2+y2+C Hence, tan-1yx=12log x2+y2+C is the required solution.
Q7.
Answer :
We have,2xydydx=x2+y2⇒dydx=x2+y22xyThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=x2+v2x22x2v⇒v+xdvdx=1+v22v⇒xdvdx=1+v22v-v⇒xdvdx=1-v22v⇒2v1-v2dv=1xdxIntegrating both sides, we get∫2v1-v2dv=∫1xdx⇒- log 1-v2=log x+log C⇒-log 1-v2x=log CPutting v=yx, we get⇒-log x2-y2x2x=log C⇒xx2-y2=C⇒x=Cx2-y2 Hence, x=Cx2-y2 is the required solution.
Q8.
Answer :
We have,x2dydx=x2-2y2+xy⇒dydx=x2-2y2+xyx2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=x2-2v2x2+x2vx2⇒v+xdvdx=1-2v2+v⇒xdvdx=1-2v2⇒11-2v2dv=1xdxIntegrating both sides, we get ∫11-2v2dv=∫1xdx⇒∫112-2v2=∫1xdx⇒122log 1+2v1-2v=log x+log C⇒log 1+2v1-2v=22log x+22 log C⇒log 1+2v1-2v=log Cx22⇒1+2v1-2v=Cx22Putting v=yx, we get⇒x+2yx-2y=Cx22Hence, x+2yx-2y=Cx22 is the required solution.
Q9.
Answer :
We have,xydydx=x2-y2⇒dydx=x2-y2xyThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=x2-v2x2vx2⇒v+xdvdx=1-v2v⇒xdvdx=1-v2v-v⇒xdvdx=1-2v2v⇒v1-2v2dv=1xdxIntegrating both sides, we get∫v1-2v2dv=∫1xdx⇒-14log 1-2v2=log x+log C⇒log 1-2v2=-4log x-4 log C⇒log 1-2v2x4=log 1C4Putting v=yx, we get⇒log x2x2-2y2=log 1C4⇒x2x2-2y2=C1whereC1=1C4Hence, x2x2-2y2=C1 is the required solution.
Q10.
Answer :
We have,y exydx=x exy+ydy⇒dxdy=x exy+yy exy⇒dxdy=xy exy+1exy⇒dxdy=xy+e-xyThis is a homogeneous differential equation.Putting x=vy and dxdy=v+ydvdy, we getv+ydvdy=v+e-v⇒ydvdy=e-v⇒evdv=1ydyIntegrating both sides, we get∫evdv=∫1ydy⇒ev=log y+CPutting v=yx, we get⇒exy=log y+CHence, exy=log y+C is the required solution.
Q11.
Answer :
x2dydx=x2+xy +y2⇒dydx=x2+xy +y2x2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=x2+x2v +v2x2x2⇒v+xdvdx=1+v+v2⇒xdvdx=1+v2⇒11+v2dv=1xdxIntegrating both sides, we get ∫11+v2dv=∫1xdx⇒tan-1 v=log x+CPutting v=yx, we get⇒tan-1 yx=log x+CHence, tan-1 yx=log x+C is the required solution.
Q12.
Answer :
We have,y2-2xy dx=x2-2xy dy⇒dydx=y2-2xyx2-2xyThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=v2x2-2vx2x2-2vx2⇒v+xdvdx=v2-2v1-2v⇒xdvdx=3v2-3v1-2v⇒1-2v3v2-3vdv=1xdxIntegrating both sides, we get∫1-2v3v2-3vdv=∫1xdx⇒-∫2v-13v2-3vdv=∫1xdx⇒-13∫6v-33v2-3vdv=∫1xdxPutting 3v2-3v=t⇒6v-3 dv=dt∴-13∫1tdt=∫1xdx⇒-13log t=log x+log CSubstituting the value of t, we get-13log 3v2-3v=log x+log C⇒-13log v2-v-13log3=log x+log C⇒-13log v2-v=log x+log C-13log3⇒-13log v2-v=log x+log C1 where, log C1=log C-13log3Substituting the value of v, we get-13log yx2-yx=log x+log C1⇒-13log y2x2-yx=log C1x⇒log y2-xyx2=-3log C1x⇒log y2-xyx2=log 1C13x3⇒y2-xyx2=1C13x3⇒xy2-x2y=1C13⇒x2y-xy2=-1C13⇒x2y-xy2=K where, log K=-1C13
Q13.
Answer :
2xy dx+x2+2y2 dy=0⇒dydx=-2xyx2+2y2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=-2vx2x2+2v2x2⇒v+xdvdx=-2v1+2v2⇒xdvdx=-2v1+2v2-v⇒xdvdx=-3v-2v31+2v2⇒1+2v23v+2v3dv=-1xdxIntegrating both sides, we get∫1+2v23v+2v3dv=-∫1xdxSubstituting 3v+2v3=t, we get31+2v2 dv=dt∴13∫dttdv=-∫1xdx⇒13log t=-log x+log C⇒13log 3v+2v3=-log x+log C⇒log 3v+2v3=-3 log x+3 log C⇒log 3v+2v3×x3 =log C3⇒3v+2v3×x3=C3Putting v=yx, we get⇒3×yx+2×y3x3×x3=C3⇒3yx2+2y3=C1Hence, 3yx2+2y3=C1 is the required solution.
Q14.
Answer :
We have,3×2 dy=3xy+y2 dx⇒dydx=3xy+y23x2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=3vx2+v2x23x2⇒v+xdvdx=3v+v23⇒xdvdx=v23⇒3v2dv=1xdxIntegrating both sides, we get3∫1v2dv=∫1xdx⇒-3×1v =log x +C⇒-3v=log x +CPutting v=yx, we get⇒-3xy=log x +CHence, -3xy=log x +C is the required solution.
Q15.
Answer :
We have,dydx=x2y+xThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=x2vx+x⇒v+xdvdx=12v+1⇒xdvdx=12v+1-v⇒xdvdx=1-2v2-v2v+1⇒2v+11-2v2-vdv=1xdxIntegrating both sides, we get ∫2v+11-2v2-vdv=∫1xdx⇒∫2v+12v2+v-1dv=-∫1xdx⇒∫2v+12vv+1-1v+1dv=-∫1xdx⇒∫2v+12v-1v+1dv=-∫1xdx …..(1)Solving left hand side integral of (1), we getUsing partial fraction,Let 2v+12v-1v+1=A2v-1+Bv+1∴ A+2B=2 …..(2) And A-B=1 …..(3) Solving (2) and (3), we get A=43 and B=13∴∫2v+12v-1v+1dv=43∫12v-1dv+13∫1v+1dv =43×2log 2v-1+13log v+1 +log C From (1), we get 23log 2v-1+13v+1 +log C =-log x+log C1⇒log 2v-12v+1=-3log x+log C2 ⇒log 2v-12v+1=log C23x3⇒2v-12v+1=C23x3Putting v=yx, we get⇒2y-xx2y+xx=C23x3⇒x+y2y-x2=k
Q16.
Answer :
x+2ydx-2x-y dy=0⇒dydx=x+2y2x-yThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=x+2vx2x-vx⇒xdvdx=1+2v2-v-v⇒xdvdx=1+v22-v⇒2-v1+v2dv=1xdxIntegrating both sides, we get∫2-v1+v2dv=∫1xdx …..(1)⇒∫21+v2dv-∫v1+v2dv=∫1xdx⇒∫21+v2dv-12∫2v1+v2dv=∫1xdx⇒2 tan-1v-12log 1+v2=log x+log C⇒2 tan-1v=log x+log C+log 1+v212⇒2 tan-1v=log Cx1+v2⇒Cx1+v2=e2 tan-1vPutting v=yx, we get⇒Cx1+yx2=e2 tan-1yx⇒Cx2+y2=e2 tan-1yxHence, x2+y2=Ke-2 tan-1yx is the required solution.
Q17.
Answer :
We have,dydx=yx-y2x2-1This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=v-v2-1⇒xdvdx=-v2-1⇒1v2-1dv=-1xdxIntegrating both sides, we get∫1v2-1dv=-∫1xdx⇒ log v+v2-1=-log x +log C⇒log v+v2-1x=log C⇒ v+v2-1x= CPutting v=yx, we get⇒yx+y2x2-1x=CHence, y+y2-x2 =C is the required solution.
Q18.
Answer :
We have,dydx=yxlog yx+1This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=vlog v+1⇒xdvdx=vlog v+v-v⇒1vlog vdv=1xdxIntegrating both sides, we get ∫1vlog vdv=∫1xdxPutting log v=t, we getdv= vdt∴ ∫1v ×t×v dt=∫1xdx⇒∫dtt=∫1xdx⇒log t=log x+log C⇒t=Cx …..(1) Substituting the value of t in (1), we get log v=CxPutting v=yx, we get⇒log yx=CxHence, log yx=Cx is the required solution.
Q19.
Answer :
dydx=yx+sin yxThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=v+ sin v⇒xdvdx=v+ sin v-v⇒1sin vdv=1xdxIntegrating both sides, we get∫1sin vdv=∫1xdx⇒∫cosec v dv=∫1xdx⇒log tan v2=log x +log C⇒log tan v2=log x+log C⇒log tan v2=log Cx⇒tan v2=CxPutting v=yx, we get⇒tan y2x=CxHence, tan y2x=Cx is the required solution.
Q20.
Answer :
We have,y2 dx+x2-xy+y2 dy=0dydx=-y2x2-xy+y2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=-v2x2x2-vx2+v2x2⇒v+xdvdx=-v21-v+v2⇒xdvdx=-v21-v+v2-v⇒xdvdx=-v-v31-v+v2⇒1-v+v2v+v3dv=-1xdx⇒1+v2-vv1+v2dv=-1xdxIntegrating both sides, we get ∫1+v2-vv1+v2dv=∫1xdx⇒∫1+v2v1+v2dv-∫vv1+v2dv=-∫1xdx⇒∫1vdv-∫11+v2dv=-∫1xdx⇒log v-tan -1v=-log x+log C⇒log vxC=tan-1v⇒vxC= etan-1vPutting v=yx, we get⇒y=Cetan-1vHence, y=Cetan-1v is the required solution.
Page 22.81 Ex. 22.9
Q21.
Answer :
We have,xx2+y2-y2dx+xy dy=0dydx=y2-xx2+y2xyThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=v2x2-xx2+v2x2vx2⇒v+xdvdx=v2-1+v2v⇒v+xdvdx=v-1+v2v⇒xdvdx=-1+v2v⇒ v1+v2dv=-1xdxPutting 1+v2=t, we getv dv=dt2∴ 12tdt=-1xdxIntegrating both sides, we get ∫ 12tdt=-∫1xdx⇒t=-log x+log C …..(1)Substituting the value of t in (1), we get1+v2=log CxPutting v=yx, we get⇒y2+x2=x log CxHence, y2+x2=x log Cx is the required solution.
Q22.
Answer :
xdydx=y-x cos2 yx⇒dydx=y-x cos2 yxx⇒dydx=yx- cos2 yxThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=v-cos2v⇒xdvdx=-cos2v⇒ 1cos2 vdv=-1xdx⇒sec2 v=-1xdxIntegrating both sides, we get ∫ sec2 v dv=-∫1xdx⇒tan v=-log x+log C ⇒tan v=log CxPutting v=yx, we gettan yx= log Cx⇒tan yx= log Cx is the required solution.
Q23.
Answer :
yxcos yxdx-xysin yx+cos yxdy=0⇒xysin yx+cos yxdy=yxcos yxdx⇒dydx=yxcos yxxysin yx+cos yxThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=v cos v1vsin v+cos v⇒xdvdx=v cos v1vsin v+cos v-v⇒xdvdx=-sin v1vsin v+cos v⇒1vsin v+cos vsin vdv=-1xdx⇒1v+cot vdv=-1xdxIntegrating both sides, we get∫1v+cot vdv=-∫1xdx⇒∫1vdv+∫ cot v dv=-∫1xdx⇒log v+log sin v=-log x+log C⇒log vxsin v=log C⇒v x sin v=CPutting v=yx, we get⇒ysin yx=CHence, ysin yx=C is the required solution.
Q24.
Answer :
Given:
xy logxy dx+y2-x2 logxydy=0⇒xy logxy dx =-y2-x2 logxydy⇒dxdy=-y2-x2 logxyxy logxy = x2 logxy-y2xy logxyIt is a homogeneous equation.We put x=vydxdy=v+ydvdySo, v+ydvdy=v2y2 log(v)-y2vy2 log(v)
v+ydvdy=v2 log(v)-1v log(v)⇒ydvdy=v2 log(v)-1v log(v)-v⇒ydvdy=v2 log(v)-1-v2 log(v)v log(v)⇒ydvdy=-1v log(v)⇒v log(v) dv=-1ydyOn integrating both sides we get,
∫v log(v) dv=-∫1ydy⇒v22 log(v)-∫v2 dv=-log y + C⇒v22 log(v) – v24=-log y + C⇒v22log(v) -12=-log y+C⇒v2 log(v) -12=-2 log y + Cnow putting back the values of v as xy we get,x2y2log(v) -12 + log y2 = C
Q25.
Answer :
We have,1+exy dx+exy 1-xy dy=0⇒dxdy=-exy 1-xy1+exyThis is a homogeneous differential equation.Putting x=vy and dxdy=v+ydvdy, we get v+ydvdy=-ev1-v1+ev⇒ydvdy=-ev1-v1+ev-v⇒ydvdy=-ev+evv-v-vev1+ev⇒ydvdy=-v+ev1+ev⇒1+evv+evdv=-1ydyIntegrating both sides, we get ∫1+evv+evdv=-∫1ydy⇒log v+ev=-log y+log C⇒v+ev=Cy⇒v+ev=CyPutting v=xy, we getxy+exy=Cy⇒x+yexy=CHence, x+yexy=C is the required solution.
Q26.
Answer :
We have,x2+y2dydx=8 x2-3xy+2y2⇒dydx=8 x2-3xy+2y2x2+y2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=8 x2-3vx2+2v2x2x2+v2x2⇒xdvdx=8-3v+2v21+v2-v⇒xdvdx=8-4v+2v2-v31+v2⇒xdvdx=42-v+v22-v1+v2⇒xdvdx=4+v22-v1+v2⇒1+v24+v22-vdv=1xdxIntegrating both sides, we get ∫1+v24+v22-vdv=∫1xdx …..(1)Let us consider the left hand side of (1).Using partial fraction,Let 1+v24+v22-v=Av+B4+v2+C2-v⇒1+v2=Av2-v+B2-v+C 4+v2⇒1+v2=2Av-Av2+2B-Bv+4C+Cv2Comparing the coefficients of both sides, we get 2A-B=0 -A+C=1 & 2B+4C=1Solving these three equations, we getA=-38, B=-34and C=58 ∴ 1+v24+v22-v=-38v-344+v2+582-v …..(2)From (1) and (2), we get∫-38v-344+v2+582-v =∫1xdx ⇒-38∫vv2+4dv-34∫1v2+4dv+58∫12-vdv=∫1xdx⇒-316log v2+4-34×2tan -1v2-58log 2-v=log x+log C⇒-34×2tan -1v2=log Cx2-v58v2+4316⇒e-38tan -1v2=Cx2-v58v2+4316Putting v=yx, we get⇒e-38tan -1y2x=Cx2-yx58yx22+4316⇒e-38tan -1y2x=Cx×1x2x-y58y2+4×2316⇒e-38tan -1y2x=C2x-y58y2+4x2316Hence, e-38tan -1y2x=C2x-y58y2+4×2316 is the required solution.
Q27.
Answer :
We have,x2-2xy dy+x2-3xy+2y2 dx=0⇒dydx=x2-3xy+2y22xy-x2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=x2-3vx2+2v2x22vx2-x2⇒v+xdvdx=1-3v+2v22v-1⇒xdvdx=1-3v+2v22v-1-v⇒xdvdx=1-2v2v-1⇒xdvdx=-1⇒dv=-1xdxIntegrating both sides, we get∫dv=-∫1xdx⇒v=-log x+CPutting v=yx, we get⇒yx=-log x+C⇒yx+log x=CHence, yx+log x=C is the required solution.
Q28.
Answer :
xdydx=y-x cos2 yx⇒dydx=y-x cos2 yxxThis is a homogeneous differential equation. Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=vx-x cos2 vx⇒v+xdvdx=v-cos2v⇒xdvdx=-cos2v⇒ sec2v dv=-1xdxIntegrating both sides, we get∫ sec2v dv=-∫1xdx ⇒tan v =- log x+log C⇒tan v =log CxPutting v=yx, we get∴ tan yx =log CxHence, tan yx =log Cx is the required solution.
Q29.
Answer :
We have,xdydx-y=2y2-x2⇒dydx=2y2-x2+yxThis is a homogeneous differential equation. Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=2v2x2-x2+vxx⇒v+xdvdx=2v2-1+v⇒xdvdx=2v2-1+v-v⇒xdvdx=2v2-1⇒12v2-1dv=1xdxIntegrating both sides, we get∫12v2-1dv=∫1xdx⇒∫1v2-1dv=2∫1xdx⇒log v+v2-1=2 log x +log C⇒v+v2-1=Cx2Putting v=yx, we get∴ yx+y2x2-1 =Cx2 ⇒y+y2-x2=Cx3Hence, y+y2-x2=Cx3 is the required solution.
Q30.
Answer :
We have,x cosyxy dx+x dy=y sin yxx dy-y dx⇒xy cos yx dx+x2cos yx dy=xy sin yx dy-y2sin yx dx⇒xy cos yx+y2sin yx dx=xy sin yx-x2cos yx dy⇒dydx=xy cos yx+y2sin yxxy sin yx-x2cos yxThis is a homogeneous differential equation. Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=vx2 cos v+v2x2sin vvx2 sin v-x2cos v⇒v+xdvdx=v cos v+v2sin vv sin v-cos v⇒xdvdx=v cos v+v2sin vv sin v-cos v-v⇒xdvdx=v cos v+v2sin v-v2 sin v+ v cos vv sin v-cos v⇒xdvdx=2v cos vv sin v-cos v⇒vsin v-cos v2 v cos vdv=1xdxIntegrating both sides, we get∫vsin v-cos v2 v cos vdv=∫1xdx⇒∫vsin v-cos vv cos vdv=2∫1xdx⇒∫v sin vv cos vdv-∫cos vv cos vdv=2∫1xdx⇒∫tan v dv-∫1vdv=2∫1xdx⇒log sec v-log v=2 log x+log C⇒log sec vv=log Cx2⇒sec vv=Cx2Putting v=yx, we getsec yx=yx×C×x2 ⇒sec yx=CxyHence, sec yx=Cxy is the required solution.
Q31.
Answer :
We have, x2+3xy+y2 dx-x2 dy=0⇒dydx=x2+3xy+y2x2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=x2+3vx2+v2x2x2⇒xdvdx=1+3v+v2-v⇒xdvdx=1+v2+2v⇒11+v2+2vdv=1xdxIntegrating both sides, we get∫11+v2+2vdv=∫1xdx⇒∫11+v2dv=∫1xdx⇒-11+v=log x+C⇒log x+11+v=-CPutting v=yx, we get∴log x+xx+y=C1 whereC1=-CHence, log x+xx+y=C1 is the required solution.
Q32
Answer :
x-y dydx=x+2y⇒dydx=x+2yx-yThis is a homogeneous differential equatiuon. Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=x+2vxx-vx⇒v+xdvdx=1+2v1-v⇒xdvdx=1+2v1-v-v⇒xdvdx=1+2v-v+v21-v⇒xdvdx=1+v+v21-v⇒1-v1+v+v2dv=1xdxIntegrating both sides, we get ∫1-v1+v+v2dv=∫1xdx
⇒∫-(v-1)v2+v+1dv=dxx⇒∫12×2v-2v2+v+1dv=∫-dxx⇒∫(2v+1)-3v2+v+1dv=-∫2dxx⇒∫(2v+1)v2+v+1dv-∫3v2+v+1dv=-∫2dxxLet I1 =∫(2v+1)v2+v+1dvand I2 =∫3v2+v+1dvI=I1 + I2
I1=∫ 2v+1v2+v+1dvlet v2+v+1=t ⇒(2v2+1)dv=dttherefore, I1=∫ 2v+1v2+v+1dv = ∫dtt=logt=logv2+v+1hence, I1 = logv2+v+1Also, I2= ∫3v2+v+1dv = ∫3v2+2v12+122-122+1=∫3v+122+322dv=323tan-1v+1232=23 tan-1v+1232I2=23 tan-1v+1232Hence, I=I1+I2=logv2+v+1 +23 tan-1v+1232therefore, logv2+v+1 +23 tan-1v+1232 =-2logx+Cputting the value of v in the above equation we get,logx2+y2+xy = 23 tan-1x+2yx3 +C
Q33.
Answer :
We have, 2x2y+y3 dx+xy2-3×3 dy=0⇒dydx=2x2y+y33x3-xy2This is a homogeneous differential equation. Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=2vx3+v3x33x3-v2x3⇒v+xdvdx=2v+v33-v2⇒xdvdx=2v+v33-v2-v⇒xdvdx=2v+v3-3v+v33-v2⇒xdvdx=2v3-v3-v2⇒3-v22v3-vdv=1xdxIntegrating both sides, we get ∫3-v22v3-vdv=∫1xdx⇒3∫12v3-vdv-∫v22v3-vdv=∫1xdx …..(1)Considering 12v3-v=1v2v2-1,let 1v2v2-1=Av+Bv+C2v2-1 …..(2)1=A2v2-1+Bv+C v⇒1=2Av2-A+Bv2+CvComparing the coeficients of both sides, we get∴ 2A+B=0 , C=0 and A=-1⇒-2+B=0⇒B=2Substituting A=-1, B=2 and C=0 in (2), we get1v2v2-1=-1v+2v2v2-1 …..(3)From (2) and (3), we get3∫-1v+2v2v2-1dv-∫v2v2-1dv=∫1xdx⇒-3∫1vdv+5∫v2v2-1dv=∫1xdx⇒-3 log v+54log 2v2-1=log x+log C⇒12 log 1v+5 log 2v2-1 4=log Cx⇒log 1v12×2v2-15=log C4x4⇒1v12×2v2-15=C4x4Putting v=yx, we get⇒x12y12×2y2x2-15=C4x4⇒2y2-x2x25=C4x4×y12x12Hence, C4x2y12=2y2-x25 is the required solution.
Q34.
Answer :
We have,xdydx-y+x sin yx=0⇒dydx=y-x sin yxxThis is a homogenoeus differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=vx-x sin vx⇒xdvdx=v-sin v-v⇒xdvdx=-sin v⇒ cosec v dv=-1xdxIntegrating both sides, we get∫cosec v dv=-∫1xdx⇒-∫cosec v dv=∫1xdx⇒-log cosec v-cot v=log x+log C⇒log 1cosec v-cot v=log Cx⇒log cosec v+cot v=log Cx⇒log 1+cos vsin v=log Cx⇒1+cos vsin v=Cx⇒x sin v=1C1+cos v⇒x sin v=K1+cos v where, K=1CPutting v=yx, we get ⇒x sinyx=K1+cosyxHence, x sinyx=K1+cosyx is the required solution.
Q35.
Answer :
We have,y dx+x log yx dy-2x dy=0⇒2x-x log yx dy=y dx⇒dydx=y2x-x log yxThis is a homogenoeus differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=vx2x-x log v⇒v+xdvdx=v2-log v⇒xdvdx=v2-log v-v⇒xdvdx=v-2v+v log v 2- log v⇒xdvdx=v log v-v2-log v⇒2-log vv log v-vdv=1xdxIntegrating both sides, we get∫2-log vv log v-vdv=∫1xdx⇒∫1-log v-1v log v-1dv=∫1xdxPutting log v-1=t⇒1vdv=dt∴∫1-ttdt=∫1xdx⇒∫1t-1dt=∫1xdx⇒log t-t=log x+log C⇒log log v-1-log v-1=log x+log C⇒log log v-1-log v=log x+log C1 where, log C1=log C-1⇒log log v-1v=log C1x⇒log v-1v=C1x⇒log v-1=C1xvPutting v=yx, we getlog yx-1=C1x×yx⇒log yx-1=C1yHence, log yx-1=C1y is the required solution.
Q36.
Answer :
(i) (x2 + y2)dx = 2xy dy, y(1) = 0
We have,
(x2 + y2) dx = 2xy …..(i)
This is a homogenous equation, so let us take y = vx
Then, dydx=v+xdvdxPutting y=vx in equation (i)x2+v2x2=2vx2v+xdvdxx21+v2=2vx2v+xdvdx1+v2=2v2+2vxdvdx1-v2=2vxdvdxdxx=2v dv1-v2
On integrating both sides, we get
∫1xdx=∫2v1-v2dvLet, 1-v2=t⇒-2v dv=dtloge x=-∫dtt loge x=-loge t+cloge x=-loge 1-y2x2+cloge xx2-y2x2=cloge x2-y2x=cAs y1=0⇒c=0∴loge x2-y2x=0⇒x2-y2x=1⇒x2-y2=x
(ii) xeyx-y+xdydx=0 ye=0
This is also a homogenous equation,
Put y = vx
dydx=v+xdvdxx ev-vx+xv+xdvdx=0x ev-vx+xv+x2dvdx=0x ev+x2dvdx=0ev=-xdvdxdxx=-1evdv
On integration both sides we get,
∫dxx=-∫1evdvloge x=-∫e-v dv⇒logex=e-yx+c ∵y=vxAs given ye=0logee=e-0e+c1=1+c⇒c=0∴logex=e-yx
(iii) dydx-yx+cosecyx=0, y1=0
This is an homogenous equation, put y = vx
dydx+v+xdvdxv+xdvdx-v+cosec v=0xdvdx=cosec vdvcosec v=dxxsin v dv=dxx
On integrating both sides, we get
∫ sin v dv=∫dxx-cos v=logex+c-cos v+logex=ccos v+logex=-ccos yx+logex=-cAs y1=0cos 01=0+loge1=-c1+0=-c⇒c=-1⇒cos yx+logex=1
(iv) (xy − y2) dx − x2 dy = 0, y(1) = 1
This is an homogenous equation, put y = vx
dydx=v+xdvdxxy-y2=x2dydxvx2-v2x2=x2v+xdvdxvx21-v=x2v+xdvdxv1-v=v+xdvdxv-v2=v+xdvdx-v2=xdvdx-1xdx=1v2dv
On integrating both sides we get,
-∫1xdx=∫1v2dv-logex=v-2+1-2+1+c-logex=v-1-1+c-logex=-1v+c-logex=-1v+cxy-logex=cAs y1=111-loge1=c⇒c=1
(v) dydx=yx+2yx2x+y, y1=1
This is an homogenous equation, put y = vx
⇒dydx=v+xdvdx
⇒v+xdvdx=vx+2vx2x+vx⇒v+xdvdx=v1+2v2+v⇒xdvdx=v1+2v-v2+v2+v⇒xdvdx=v+2v2-2v-v22+v⇒ xdvdx=v2-v2+v⇒2+vdvv2-v=dxx
On integrating both side of the equation we get,
∫2+vv2-vdv=∫dxx⇒∫2vv-1dv+∫vvv-1dv=∫dxx⇒2∫11-vdv-∫1vdv+∫1v-1dv=logex+c⇒2logev-1-logev+logev-1=logex+c2loge v-1v+logev-1=logex+c2logey-xy+logey-xx=logex+cAs y1=22 loge2-12+loge2-11=loge1+c2 loge12+loge1=loge1+c-2 loge2+0=0+c-2 loge2=c∴2 logey-xy+logey-xx=logex-2 loge2
(vi) (y4 − 2x3 y) dx + (x4 − 2xy3) dy = 0, y (1) = 1
This is an homogenous equation, put y= vx
v4x4-2vx4+x4-2v3x4 v+xdvdx=0v4x4-2vx4=2v3x4-x4 v+xdvdxvx4v3-2=x42v3-1 v+xdvdxvv3-2=2v3-1v+x2v3-1dvdxvv3-2-2v3+1=x2v3-1dvdxv-1-v3=x2v3-1dvdxv1+v3=x1-2v3dvdxdxx=1-2v3v1+v3dv
On integrating both side of the equation we get,
∫dxx=∫1-2v3v1+v3dv⇒logex=∫1+v3-3v3v1+v3dv⇒logex=∫1+v3v1+v3dv-∫3vv1+v3dv⇒logex=∫1vdv-∫3v21+v3dv⇒logex=logev-∫dtt⇒logex=logev-loge1+v3+c let 1+v3=t, 3v2dv=dt⇒logex=logev1+v3+cAs v=yx⇒logex=logeyx1+y3x+c⇒logex=logeyx2x3+y3+cAs y1=1⇒loge1=loge11+1+c⇒0=loge12+cc=-loge12⇒c=loge2∴logex=logeyx2x3+y3+loge2
(vii) x(x2+3y2)dx +y(y2+3×2)dy = 0, y(1) = 1 dydx=-x(x2+3y2)y(y2+3×2)it is a homogeneous equation. Put y=vxand dydx=v+xdvdx
So, v+xdvdx =-x(x2+3v2x2)vx(v2x2+3×2)xdvdx=-(1+3v2)v(v2+3)-3= -1-3v2-v4-3v2v(v2+3)xdvdx=-v4-6v2-1v(v2+3)v(v2+3)v4+6v2+1dv=-dxx∫4v3+12vv4+6v2+1dv=-4∫dxxlogv4+6v2+1=logcx4v4+6v2+1=cx4y4+6y2x2+x4=c ….(1)
put y=1, x=1(1+6+1)=c ⇒c=8put c=8 in equation (1),(y4+x4+6x2y2)=8
(viii) {x sin2yx-y}dx + x dy = 0, y(1) = π4it is a homogeneous equation. so, we put y= vxdydx=v+xdvdxso, v+xdvdx=-sin2vxx+vxxxdvdx=-sin2vdvsin2v=-dxxintegrating both sides, we getcotyx=logcx
putting the values of x=1 and y=π4cotπ4=log c1=log cc=eHence, cotyx=log(ex)
(ix) xdydx-y+x sinyx=0, y(2)=πit is a homogeneous equation. put y=vxand dydx=v+xdvdxso, v+xdvdx=vxx-sinvxxxdvdx=-sinvdvsinv=-dxxcosec(v)dv=-dxxintegraing both sides we get,log(cosec(v)-cot(v))=-log x + log c
logcosecyx-cotyx=-log x + log cputting the values x=2 and y=π logcosecπ2-cotπ2=-log 2 + log cc=0logcosecyx-cotyx=-log x
Q37.
Answer :
x cos yxdydx=y cos yx+x⇒dydx=y cos yx+xx cos yxThis is a homogeneous differential equation. Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=vx cos v+xx cos v⇒v+xdvdx=v cos v+1cos v⇒xdvdx=v cos v+1-v cos vcos v⇒xdvdx=1cos v⇒cos v dv=1xdxIntegrating both sides, we get∫cos v dv=∫1xdx⇒sin v=log x+CPutting v=yx, we getsinyx=log x+C …..1At x=1, y=π4 GivenPutting x=1 and y=π4 in (1), we getC=12Putting C=12 in (1), we getsin yx=log x+12Hence, sin yx=log x+12 is the required solution.
Q38.
Answer :
x-ydydx=x+2y⇒dydx=x+2yx-yThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=x+2vxx-vx⇒xdvdx=1+2v1-v-v⇒xdvdx=1+2v-v+v21-v⇒xdvdx=1+v+v21-v⇒1-v1+v+v2 dv=1xdxIntegrating both sides, we get ∫1-v1+v+v2 dv=∫1xdx⇒∫11+v+v2dv-12∫2v+1-11+v+v2=∫1xdx⇒∫11+v+v2dv-12∫2v+11+v+v2dv+12∫11+v+v2dv=∫1xdx ⇒32∫11+v+v2dv-12∫2v+11+v+v2dv=∫1xdx ⇒32∫11+v+v2+14-14dv-12∫2v+11+v+v2dv=∫1xdx ⇒32∫1v+122+322dv-12∫2v+11+v+v2dv=∫1xdx⇒3tan -1 v+1232-12log 1+v+v2=log x+CPutting v=yx, we get3tan-12y+x3x -12log x2+xy+y2x2=log x+C⇒3tan-12y+x3x -12log x2+xy+y2+log x=log x+C ⇒3tan-12y+x3x -12log x2+xy+y2=C …..(1) At x=1, y=0 GivenPutting x=1 and y=0 in (1), we get3 tan-113 -12log 1=C⇒C=3 tan-113⇒C=3 ×π6⇒C=π23Substituting the value of C in (1), we get 3tan-12y+x3x -12log x2+xy+y2=π23⇒23tan-12y+x3x -log x2+xy+y2=π3⇒log x2+xy+y2=23tan-12y+x3x-π3Hence, log x2+xy+y2=23tan-12y+x3x-π3 is the required solution.
Page 22.102 Ex. 22.10
Q1.
Answer :
We have,dydx+2y=e3x …..(1)Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2Q=e3x∴ I.F.=e∫P dx =e∫2 dx = e2xMultiplying both sides of (1) by e2x, we gete2x dydx+2y=e2xe3x⇒e2xdydx+2e2xy=e5xIntegrating both sides with respect to x, we gety e2x=∫e5xdx+C⇒y e2x=e5x5+C⇒y=15e3x+Ce-2xHence, y=15e3x+Ce-2x is the required solution.
Q2.
Answer :
We have,
4dydx+8y=5 e-3x
⇒dydx+2y=54e-3x …..(1)Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2Q=54e-3x∴ I.F.=e∫P dx =e∫2 dx = e2xMultiplying both sides of (1) by e2x, we gete2x dydx+2y=54e2xe-3x⇒e2xdydx+2e2xy=54e-xIntegrating both sides with respect to x, we gety e2x=54∫e-x dx+C⇒y e2x=-54e-x+C⇒y=54e-3x+Ce-2xHence, y=54e-3x+Ce-2x is the required solution.
Q3.
Answer :
We have, dydx+2y=6ex …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2Q=6ex∴ I.F.=e∫P dx =e∫2 dx = e2xMultiplying both sides of 1 by e2x, we gete2x dydx+2y=6e2xex⇒e2xdydx+2e2xy=6e3xIntegrating both sides with respect to x, we gety e2x=6∫e3xdx+C⇒y e2x=6e3x3+C⇒y e2x=2e3x+CHence, y e2x=2e3x+C is the required solution.
Q4.
Answer :
We have, dydx+y=e-2x …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1Q=e-2x ∴ I.F.=e∫P dx =e∫1 dx = exMultiplying both sides of 1 by ex, we getex dydx+y=exe-2x⇒exdydx+exy=e-xIntegrating both sides with respect to x, we gety ex=∫e-xdx+C⇒y ex=-e-x+C⇒y=-e-2x+Ce-xHence, y=-e-2x+Ce-x is the required solution.
Q5.
Answer :
We have,xdydx=x+y⇒dydx=1+1xy ⇒dydx-1xy=1 …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1x Q=1∴ I.F.=e∫P dx =e-∫1x dx =e-log x =elog 1x =1xMultiplying both sides of 1 by 1x, we get1x dydx-1xy=1x×1⇒1xdydx-1x2y=1xIntegrating both sides with respect to x, we gety1x=∫1x dx+C⇒yx=log x+CHence, yx=log x+C is the required solution.
Q6.
Answer :
We have,dydx+2y=4x …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2 Q=4x ∴ I.F.=e∫P dx =e∫2 dx = e2xMultiplying both sides of 1 by e2x, we gete2x dydx+2y=e2x4x ⇒e2xdydx+2e2xy=e2x4x Integrating both sides with respect to x, we gety e2x=4∫x e2x dx+C⇒y e2x=4∫xI e2xII dx+C⇒y e2x=4x∫e2x dx-4∫ddxx∫e2x dxdx+C⇒y e2x=4xe2x2-4×12∫e2x dx+C⇒y e2x=2x e2x-4×14e2x+C⇒y e2x=2x e2x-e2x+C⇒y e2x=2x-1e2x+C⇒y=2x-1+Ce-2xHence, y=2x-1+Ce-2x is the required solution.
Q7.
Answer :
We have,xdydx+y=x ex⇒dydx+1xy=ex …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1x Q=ex∴ I.F.=e∫P dx =e∫1x dx =elog x =x Multiplying both sides of 1 by x, we getxdydx+1xy=x ex⇒xdydx+y=xexIntegrating both sides with respect to x, we getxy=∫x exdx+C⇒xy=∫xI exII dx+C⇒xy=x∫ex dx-∫ddxx∫ex dxdx+C⇒xy=x ex-ex+C⇒xy=x-1ex+C⇒y=x-1xex+CxHence, y=x-1xex+Cx is the required solution.
Q8.
Answer :
We have, dydx+4xx2+1y+1×2+12=0 ⇒dydx+4xx2+1y=-1×2+12 …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=4xx2+1 Q=-1×2+12∴ I.F.=e∫P dx =e2∫2xx2+1 dx =e2log x2+1 =x2+12Multiplying both sides of 1 by x2+12, we getx2+12dydx+4xx2+1y=x2+12-1×2+12 ⇒x2+12dydx+4xx2+1y=-1Integrating both sides with respect to x, we getx2+12y=-∫dx+C⇒x2+12y=-x+CHence, x2+12y=-x+C is the required solution.
Q9.
Answer :
We have,
xdydx+y=x log x
Dividing both sides by x, we get
dydx+yx=log xComparing with dydx+Py=Q, we getP=1xQ=log xNow, I.F.=e∫Pdx=e∫1xdx =elogx =xSo, the solution is given byy×I.F.=∫Q×I.F. dx+C⇒xy=∫x IIlog xI dx+C⇒xy=log x∫xdx-∫ddxlog x∫x dxdx+C⇒xy=x2 log x2-∫x2dx+C⇒xy=x2 log x2-x24+C⇒4xy=2 x2log x-x2+K where, K=2C
Q10.
Answer :
We have, xdydx-y=x-1ex⇒dydx-1xy=x-1xex …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1x Q=x-1xex∴ I.F.=e∫P dx =e-∫1x dx =e-log x =1xMultiplying both sides of 1 by I.F.=1x, we get1x dydx-1xy=1xx-1xex ⇒1xdydx-1x2y=x-1x2exIntegrating both sides with respect to x, we get1xy=∫1x-1x2ex dx+C⇒1xy=exx+C⇒y=ex+CxHence, y=ex+Cx is the required solution.
Q11.
Answer :
We have, dydx+yx=x3 …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1xQ=x3∴ I.F.=e∫P dx =e∫1x dx =elog x =x Multiplying both sides of 1 by x, we getx dydx+1xy=x x3 ⇒xdydx+y=x4Integrating both sides with respect to x, we getxy=∫x4 dx+C⇒xy=x55+C⇒5xy=x5+5C⇒5xy=x5+K where, K=5CHence, 5xy=x5+K is the required solution.
Q12.
Answer :
We have,dydx+y=sin x …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1Q=sin x ∴ I.F.=e∫P dx =e∫ dx = exMultiplying both sides of 1 by ex, we getex dydx+y=exsin x⇒exdydx+exy=exsin x Integrating both sides with respect to x, we gety ex=∫exsin x dx+C⇒y ex=ex2sin x-cos x+C⇒y=Ce-x+12sin x-cos xHence, y=Ce-x+12sin x-cos x is the required solution.
Q13.
Answer :
We have,dydx+y=cos x …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1Q=cos x ∴ I.F.=e∫P dx =e∫ dx = exMultiplying both sides of (1) by ex, we getex dydx+y=excos x ⇒exdydx+exy=excos xIntegrating both sides with respect to x, we gety ex=∫excos x dx+C⇒y ex=12∫excos x+sin x+-sin x+cos x dx+C⇒yex=ex2cos x+sin x+C⇒y=12cos x+sin x+Ce-xHence, y=12cos x+sin x+Ce-x is the required solution.
Q14.
Answer :
We have,dydx+2y=sin x …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2andQ=sin x ∴ I.F.=e∫P dx =e∫2 dx = e2xMultiplying both sides of 1 by I.F.=e2x, we gete2x dydx+2y=e2xsin x ⇒e2xdydx+2e2xy=e2xsin xIntegrating both sides with respect to x, we gety e2x=∫e2xsin x dx+C⇒y e2x=15∫2e2x2sin x-cos x+e2x2 cos x+sin x dx+CPutting e2x2 sin x-cos x=t⇒2e2x2sin x-cos x+e2x2 cos x+sin x dx=dt∴y e2x=15∫dt+C⇒y e2x=t5+Cy e2x=e2x52sin x-cos x+C⇒y=152sin x-cos x+Ce-2xHence, y=152sin x-cos x+Ce-2x is the required solution.
Q15.
Answer :
We have, dydx=y tan x-2sin xdydx-y tan x=-2sin x …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-tan xQ=-2sin x ∴ I.F.=e∫P dx =e-∫tan x dx = e-logsec x=cos xMultiplying both sides of 1 by cos x, we getcos x dydx-y tan x=-2sin x×cos x⇒cos xdydx-ysin x=-sin 2x Integrating both sides with respect to x, we gety cos x=-∫sin 2x dx+C⇒ycos x=cos 2×2+C⇒2y cos x=cos 2x+2C⇒2y cos x=cos 2x+K, where k=2CHence, 2y cos x=cos 2x+K is the required solution.
Q16.
Answer :
We have, 1+x2dydx+y=tan-1x⇒dydx+y1+x2=tan-1×1+x2 …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=11+x2 Q=tan-1×1+x2∴ I.F.=e∫P dx =e∫11+x2 dx =etan-1xMultiplying both sides of 1 by etan-1x, we getetan-1x dydx+y1+x2=etan-1x tan-1×1+x2⇒etan-1xdydx+etan-1xy1+x2=etan-1xtan-1×1+x2Integrating both sides with respect to x, we getetan-1xy=∫tan-1x×etan-1×1+x2 dx+C⇒etan-1xy=I+C …..2Here, I=∫tan-1x×etan-1×1+x2 dxPutting tan-1 x=t, we get⇒11+x2dx=dt∴ I=∫t et dt =t∫etdt-∫ddtt∫etdtdt =t et-et =t-1et =tan-1x-1etan-1xSubstituting the value of I in 2, we getetan-1xy=tan-1x-1etan-1x+C⇒y=tan-1x-1+Ce-tan-1xHence, y=tan-1x-1+Ce-tan-1x is the required solution.
Q17.
Answer :
We have,dydx+y tan x=cos x …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=tan xQ=cos x ∴ I.F.=e∫P dx =e∫tan x dx = elogsec x=sec xMultiplying both sides of 1 by sec x, we getsec xdydx+y tan x=cos x ×sec x⇒sec xdydx+y sec x tan x=1Integrating both sides with respect to x, we gety sec x=∫dx+C⇒y sec x=x+CHence, y sec x=x+C is the required solution.
Q18.
Answer :
We have,dydx+y cot x=x2cot x+2x …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=cot xQ=x2cot x+2x ∴ I.F.=e∫P dx =e∫cot x dx = elogsin x=sin xMultiplying both sides of 1 by sin x, we getsin xdydx+ycot x=sin xx2cot x+2x⇒sin xdydx+ycos x=x2cos x+2x sin x Integrating both sides with respect to x, we gety sin x=∫x2Icos xIIdx+∫2x sin x dx+C⇒y sin x=x2∫cos xdx-∫ddxx2∫cos x dxdx+∫2xsin x dx+C⇒y sin x=x2sin x-∫2xsin x dx+∫2xsin x dx+C⇒y sin x=x2sin x+CHence, y sin x=x2sin x+C is the required solution.
Q19.
Answer :
We have,
dydx+y tan x=x2 cos2 x
Comparing with dydx+Py=Q, we getP=tan x Q=x2 cos2 xNow,I.F.=e∫tan x dx =elog sec x=sec xTherefore, solution is given byy×I.F.=∫x2 cos2 x×I.F. dx+C⇒y sec x=∫x2 cos x dx+C⇒y sec x=I+CWhere, I=∫x2IIcos x Idx+C⇒ I=x2∫cos x dx-∫ddxx2∫cos x dxdx⇒ I=x2sin x-2∫x sin x dx⇒ I=x2sin x-2∫xI sin xII dx⇒ I=x2sin x-2x∫sin x dx+2∫ddxx∫sin x dxdx⇒ I=x2sin x+2x cos x -2∫cos x dx⇒ I=x2sin x+2x cos x -2sin x⇒ I=x2sin x+2x cos x -2sin x∴ y sec x=x2sin x+2x cos x-2sin x+C⇒y sec x=x2sin x+2x cos x-2sin x+C
Q20.
Answer :
We have, 1+x2dydx+y=etan-1x⇒dydx+y1+x2=etan-1 x1+x2 …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=11+x2Q=etan-1 x1+x2∴ I.F.=e∫P dx =e∫11+x2 dx =etan-1xMultiplying both sides of 1 by etan-1x, we getetan-1x dydx+y1+x2=etan-1xetan-1 x1+x2⇒etan-1xdydx+y etan-1×1+x2=etan-1xetan-1×1+x2Integrating both sides with respect to x, we gety etan-1x=∫e2tan-1×1+x2 dx+C⇒y etan-1x=I+C …..2Here,I=∫e2tan-1×1+x2 dxPutting tan-1 x=t, we get11+x2dx=dt∴ I=∫e2t dt =e2t2 =e2tan-1x2Putting the value of I in 2, we gety etan-1x=e2tan-1×2+C⇒2y etan-1x=e2tan-1x+2C⇒2y etan-1x=e2tan-1x+k, where k=2CHence, 2y etan-1x=e2tan-1x+k is the required solution.
Q21.
Answer :
We have, x dy=2y+2×4+x2dx⇒dydx=2xy+2×3+x⇒dydx-2xy=2×3+x …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-2xQ=2×3+x∴ I.F.=e∫P dx =e-∫2x dx =e-2log x =1x2Multiplying both sides of 1 by 1×2, we get1x2 dydx-2xy=1×2 2×3+x⇒1x2dydx-2x3y=2x+1xIntegrating both sides with respect to x, we get1x2y=∫2x+1xdx+C⇒1x2y=x2+log x+C⇒y=x4+x2log x+Cx2Hence, y=x4+x2log x+Cx2 is the required solution.
Q22.
Answer :
We have,1+y2+x-etan-1ydydx=0⇒x-etan-1ydydx=-1+y2⇒dydx=-1+y2x-etan-1y⇒dxdy=-x-etan-1y1+y2⇒dxdy+x1+y2=etan-1 y1+y2 …..1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=11+y2Q=etan-1 y1+y2 ∴ I.F.=e∫P dy =e∫11+y2 dy =etan-1yMultiplying both sides of 1 by etan-1y, we getetan-1y dxdy+x1+y2=etan-1y etan-1 y1+y2⇒etan-1ydxdy+x etan-1×1+y2=e2tan-1 y1+y2Integrating both sides with respect to y, we getx etan-1y=∫e2tan-1 y1+y2 dy+C⇒x etan-1y=I+C …..2Here,I=∫e2tan-1 y1+y2 dyPutting tan-1 y=t, we get11+y2dy=dt∴ I=∫e2t dt =e2t2 =e2tan-1y2Putting the value of I in 2, we getx etan-1y=e2tan-1y2+C⇒2x etan-1y=e2tan-1y+2C⇒2x etan-1y=e2tan-1y+k where k=2CHence, 2x etan-1y=e2tan-1y+k is the required solution.
Q23.
Answer :
We have,y2dxdy+x-1y=0⇒y2dxdy+x=1y ⇒dxdy+1y2x=1y3 …..1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=1y2Q=1y3∴ I.F.=e∫P dy =e∫1y2dy = e-1yMultiplying both sides of 1 by e-1y, we get e-1ydxdy+x1y2= e-1y1y3⇒ e-1ydxdy+x1y2 e-1y= e-1y 1y3Integrating both sides with respect to y, we getx e-1y=∫ e-1y1y3dy+C⇒x e-1y=I+C …..2whereI=∫ e-1y1y3dyPutting t=1y, we getdt=-1y2dy∴ I=-∫ t Ie-tIIdt =-t∫e-tdt+∫ddtt∫e-tdtdt =te-t+e-t =t+1e-t =1y+1e-1yPutting the value of I in 2, we getx e-1y=1y+1e-1y+C ⇒x=y+1y+Ce1yHence, x=y+1y+Ce1y is the required solution.
Q24.
Answer :
We have, 2x-10y3dydx+y=0⇒2x-10y3dydx=-y ⇒dxdy=-1y2x-10y3 ⇒dxdy+2yx=10y2 …..1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=2yQ=10y2∴ I.F.=e∫P dy =e∫2ydy = e2log y=y2Multiplying both sides of 1 by y2, we get y2dxdy+2yx= y2×10y2⇒ y2dxdy+2yxy2=10y4Integrating both sides with respect to y, we getx y2=∫ 10y4 dy+C⇒xy2=2y5+C⇒x=2y3+Cy-2Hence, x=2y3+Cy-2 is the required solution.
Q25.
Answer :
We have,x+tan ydy=sin 2y dx⇒dxdy=x cosec 2y+12sec2y ⇒dxdy-x cosec 2y=12sec2y …..1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=-cosec 2yQ=12sec2y∴ I.F.=e∫P dy =e-∫cosec 2y dy = e-12logtan y=1tan yMultiplying both sides of 1 by 1tan y, we get 1tan ydxdy-x cosec 2y=12 1tan y×sec2y⇒ 1tan ydxdy-x cosec 2y1tan y=12 1tan y×sec2y Integrating both sides with respect to y, we get1tan yx=∫ 12 1tan y×sec2y dy+C⇒xtan y=I+C …..2where I=∫12 1tan y×sec2y dyPutting t=tan y, we getdt=sec2 y dy∴ I=12∫1t×dt =t =tan yPutting the value of I in 2, we getxtan y=tan y+C ⇒x=tan y+Ctan yHence, x=tan y+Ctan y is the required solution.
Q26.
Answer :
We have, dx+x dy=e-ysec2y dy⇒ dx=e-ysec2y dy-x dy⇒ dxdy=e-ysec2y-x⇒ dxdy+x=e-ysec2y …..1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=1Q=e-ysec2y∴ I.F.=e∫P dy =e∫dy =eyMultiplying both sides of 1 by ey, we getey dxdy+x=ey e-ysec2y⇒eydxdy+x ey=sec2yIntegrating both sides with respect to y, we getx ey=∫sec2y dy+C⇒x ey=tan y+CHence, x ey=tan y+C is the required solution.
Q27.
Answer :
We have,dydx=y tan x-2sin xdydx-y tan x=-2sin x …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-tan xQ=-2sin x∴ I.F.=e∫P dx =e-∫tan x dx = e-logsec x=cos xMultiplying both sides of 1 by cos x, we getcos x dydx-ytan x=-2sin x ×cos x⇒cos xdydx+ysin x=-sin 2xIntegrating both sides with respect to x, we gety cos x=-∫sin 2x dx+C⇒y cos x=cos 2×2+C⇒y cos x=1-2sin2x2+C⇒y cos x=-sin2x+12+C⇒y cos x=-sin2x+K where k=12+C⇒y=sec x-sin2x+KHence, y=sec x-sin2x+K is the required solution.
Q28.
Answer :
We have, dydx+y cos x=sin x cos x …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=cos xQ=sin x cos x∴ I.F.=e∫P dx =e∫cosx dx = esin xMultiplying both sides of 1 by esin x, we get esin xdydx+y cos x= esin xsin x cos x⇒ esin xdydx+esin xy cos x= esin x sin xcos x Integrating both sides with respect to x, we gety esin x=∫ esin x sin xcos x dx+C⇒y esin x=I+C …..2whereI=∫esin x sin x cos x dxPutting t=sin x, we getdt=cos x dx∴ I=∫etII tI dt =t∫etdt-∫ddtt∫etdtdt =t et-et =ett-1 = esin xsin x-1Putting the value of I in 2, we gety esin x= esin xsin x-1+C⇒y=sin x-1+Ce-sin x Hence, y=sin x-1+Ce-sin x is the required solution.
Page 22.103 Ex. 22.10
Q29.
Answer :
We have,1+x2dydx-2xy=x2+2×2+1⇒dydx-2xyx2+1=x2+2 …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-2xx2+1Q=x2+2∴ I.F.=e∫P dx =e-∫2xx2+1 dx =e-logx2+1 =1×2+1Multiplying both sides of 1 by 1×2+1, we get1x2+1 dydx-2xyx2+1=1×2+1×2+2⇒1×2+1dydx-2xyx2+12=x2+2×2+1Integrating both sides with respect to x, we get1x2+1y=∫x2+2×2+1 dx+C⇒yx2+1=∫x2+1+1×2+1 dx+C⇒yx2+1=∫dx+∫1×2+1 dx+C⇒1×2+1y=x+tan-1x +C⇒y=x+tan-1x +Cx2+1Hence, y=x+tan-1x +Cx2+1 is the required sol
Q30.
Answer :
We have,sin xdydx+y cos x=2 sin2 x cos x⇒dydx+y cot x=2sin x cos x …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=cot xQ=2sin x cos x∴ I.F.=e∫P dx =e∫cot x dx = elogsin x=sin xMultiplying both sides of 1 by sin x, we getsin xdydx+y cot x=sin x×2sin xcos x⇒sin xdydx+y cosx=2sin2 xcos xIntegrating both sides with respect to x, we gety sin x=2∫sin2 x cos x dx+C …..2Putting sin x=t⇒cos x dx=dtTherefore, 2 becomesy sin x=2∫t2 dt+C⇒y sin x=23t3+C⇒y sin x=23sin3x+CHence, y sin x=23sin3x+C is the required solution.
Q31.
Answer :
We have,x2-1dydx+2x+2y=2x+1⇒dydx+2x+2×2-1y=2x+1×2-1 ⇒dydx+2x+2×2-1y=2x-1 …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2x+2×2-1Q=2x-1∴ I.F.=e∫P dx =e∫2x+2×2-1 dx =e∫2xx2-1+4∫ 1×2-1dx =elogx2-1+4×12logx-1x+1 =elogx2-1×x-12x+12 =elogx-13x+1 =x-13x+1Multiplying both sides of 1 by x-13x+1, we getx-13x+1 dydx+2x+2×2-1y=x-13x+1×2x-1⇒x-13x+1dydx-2x+2x-12x+12y=2x-12x+1Integrating both sides with respect to x, we getx-13x+1y=∫2x-12x+1 dx+C⇒x-13x+1y=∫2x+12-4xx+1 dx+C⇒x-13x+1y=∫2x+1-4xx+1 dx+C⇒x-13x+1y=∫2x+1-4x+1-1x+1 dx+C⇒x-13x+1y=∫2x+1-4+4x+1 dx+C⇒x-13x+1y=∫2x-6+8x+1 dx+C⇒x-13x+1y=x2-6x+8log x+1+C⇒y=x+1x-13×2-6x+8log x+1+CHence, y=x+1x-13×2-6x+8logx+1 +C is the required solution.
Q32.
Answer :
We have, xdydx+2y=x cos x⇒dydx+2xy=cos xComparing with dydx+Py=Q, we getP=2xQ=cos xNow, I.F.=e∫2xdx =e2log x=x2Solution is given by,y×I.F.=∫cos x×I.F. dx+C⇒yx2=∫x2 cos x dx+C⇒x2y=I+C …..1Where,I=∫x2IIcos x Idx+C⇒ I=x2∫cos x dx-∫ddxx2∫cos x dxdx⇒ I=x2sin x-2∫x sin x dx⇒ I=x2sin x-2∫xI sin xII dx⇒ I=x2sin x-2x∫sin x dx+2∫ddxx∫sin x dxdx⇒ I=x2sin x+2x cos x -2∫cos x dx⇒ I=x2sin x+2x cos x -2sin x⇒ I=x2sin x+2x cos x -2sin xTherefore 1 becomes∴x2y=x2sin x+2x cos x-2sin x+C
Q33.
Answer :
We have, dydx-y=x ex …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1 Q=ex∴ I.F.=e∫P dx =e-∫dx =e-xMultiplying both sides of 1 by e-x, we gete-xdydx-y=x exe-x⇒e-xdydx-e-xy=xIntegrating both sides with respect to x, we gete-xy=∫x dx+C⇒e-xy=x22+C⇒y=x22+Ce xHence, y=x22+Ce x is the required solution.
Q34.
Answer :
We have, dydx+2y=xe4x …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2Q=xe4x∴ I.F.=e∫P dx =e∫2dx = e2xMultiplying both sides of 1 by e2x, we get e2xdydx+2y= e2x×xe4x⇒ e2xdydx+ 2e2xy= xe6xIntegrating both sides with respect to x, we get e2xy=∫e6xII xI dx+C⇒e2xy=x∫e6xdx-∫ddxx∫e6xdxdx+C⇒e2xy=xe6x6-e6x36+C⇒y=xe4x6-e4x36+Ce-2xHence, y=xe4x6-e4x36+Ce-2x is the required solution.
Q35.
Answer :
We have, x+2y2dydx=y⇒dxdy=1yx+2y2 ⇒dxdy-1yx=2y …..1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=-1yQ=2y∴ I.F.=e∫P dy =e-∫1ydy = e-log y=1yMultiplying both sides of 1 by 1y, we get1ydxdy-1yx= 1y×2y⇒1ydxdy-1y2x=2Integrating both sides with respect to y, we getx1y=∫ 2dy+C⇒x1y=2y+C⇒x=2y2+Cy …..2Now,y=1 at x=2∴ 2=2+C⇒C=0Putting the value of C in 2, we getx=2y2Hence, x=2y2 is the required solution.
Q36.
Answer :
i) We have,dydx+3y=emx …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=3 Q=emx∴ I.F.=e∫P dx =e∫3 dx = e3xMultiplying both sides of (1) by e3x, we get e3x dydx+3y=e3xemx ⇒ e3xdydx+3 e3xy=em+3xIntegrating both sides with respect to x, we getye3x=∫em+3x dx+C when m+3≠0 ⇒ye3x=em+3xm+3+C⇒y=emxm+3+Ce-3x ye3x=∫e0×x dx+C when m+3=0 ⇒ye3x=∫dx+C⇒ye3x=x+C⇒y=x+Ce-3xHence, y=emxm+3+Ce-3x, where m+3≠0 and y=x+Ce-3x, where m+3=0 are required solutions.
ii) We have,dydx-y=cos 2x …..(1)Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1Q=cos 2x∴ I.F.=e∫P dx =e-∫ dx = e-xMultiplying both sides of (1) by e-x, we gete-x dydx-y=e-xcos 2x ⇒e-xdydx-e-xy=e-xcos 2xIntegrating both sides with respect to x, we gety e-x=∫e-xcos 2x dx+C ⇒y e-x=I+C …..(2)Where,I=∫e-xcos 2x dx …..(3)⇒I=12e-xsin 2x-12∫-e-xsin 2x dx⇒I=12e-xsin 2x+12∫e-xsin 2x dx⇒I=12e-xsin 2x-14e-xcos 2x-12×12∫-e-x×-cos 2x dx⇒I=12e-xsin 2x-14e-xcos 2x-14∫e-xcos 2x dx⇒I=12e-xsin 2x-14e-xcos 2x-14I From 3⇒54I=12e-xsin 2x-14e-xcos 2x⇒5I=2e-xsin 2x-e-xcos 2x⇒I=e-x52sin 2x-cos 2x …..(4)From (2) and (4) we get⇒ye-x=e-x52sin 2x-cos 2x+C⇒y=152sin 2x-cos 2x+CexHence, y=152sin 2x-cos 2x+Cex is the required solution.
iii) We have, xdydx-y=x+1e-x⇒dydx-1xy=x+1xe-x …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1xQ=x+1xe-x∴ I.F.=e∫P dx =e-∫1x dx =e-log x =1xMultiplying both sides of 1 by 1x, we get1x dydx-1xy=1xx+1xe-x ⇒1xdydx-1x2y=x+1x2e-xIntegrating both sides with respect to x, we get1xy=∫1x+1x2e-xdx+C …..2Putting 1xe-x=t⇒-1xe-x-1x2e-xdx=dt⇒1x+1x2e-xdx=-dtTherefore 2 becomes1xy=-∫dt+C⇒1xy=-t+C⇒1xy=-1xe-x+C⇒y=-e-x+CxHence, y=-e-x+Cx is the required solution.
iv) We have, xdydx+y=x4⇒dydx+1xy=x3 …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1x Q=x3∴ I.F.=e∫P dx =e∫1x dx =elog x =xMultiplying both sides of 1 by x, we getx dydx+1xy=x.x3 ⇒xdydx+y=x4 Integrating both sides with respect to x, we getxy=∫x4 dx+C⇒xy=x55+C⇒y=x45+CxHence, y=x45+Cx is the required solution.
v) We have,x log xdydx+y=log xDividing both sides by x log x, we getdydx+yx log x= log xx log x⇒dydx+yx log x=1 x⇒dydx+1x log xy=1 xComparing with dydx+Py=Q, we getP=1x log x Q=1 xNow,I.F.=e∫Pdx=e∫1x log xdx =eloglog x =log xSo, the solution is given byy×I.F.=∫Q×I.F. dx+C⇒y log x=∫1 x×log x dx+C⇒y log x=log x22+C⇒y=12log x+Clog x
vi) We have, dydx-2xy1+x2=x2+2 …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-2×1+x2 Q=x2+2∴ I.F.=e∫P dx =e-∫2×1+x2 dx =e-log1+x2 =11+x2Multiplying both sides of 1 by 11+x2, we get11+x2 dydx-2xy1+x2=11+x2x2+2⇒11+x2dydx-2xy1+x22=x2+2×2+1Integrating both sides with respect to x, we get11+x2y=∫x2+2×2+1 dx+C⇒11+x2y=∫x2+1+1×2+1 dx+C⇒11+x2y=∫dx+∫1×2+1 dx+C⇒11+x2y=x+tan-1x +C⇒y=1+x2x+tan-1x +CHence, y=1+x2x+tan-1x +C is the required solution.
vii) We have,dydx+y cos x=esin x cos x …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=cos x Q=esin x cos x∴ I.F.=e∫P dx =e∫cos x dx =esin xMultiplying both sides of 1 by esin x, we getesin x dydx+y cos x=esin x×esin x cos x⇒esin xdydx+yesin xcos x=e2sin x cos xIntegrating both sides with respect to x, we getesin xy=∫e2sin x cos x dx+C⇒esin xy=I+C …..2Where,I=∫e2sin x cos x dxPutting t=sinx, we getdt=cos x dx∴ I=∫e2t dt =e2t2 =e2sin x2Putting the value of I in 2, we getesin xy=e2sin x2+C⇒y=esin x2+Ce-sin xHence, y=esin x2+Ce-sin x is the required solution.
viii) We have,x+ydydx=1⇒dydx=1x+y⇒dxdy=x+y⇒dxdy-x=y …..1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=-1Q=y∴ I.F.=e∫P dy =e∫-1 dy =e-yMultiplying both sides of (1) by e-y, we gete-y dxdy-x=e-yy⇒e-ydxdy-e-yx=e-yyIntegrating both sides with respect to y, we gete-yx=∫y Ie-yII dy+C⇒e-yx=y∫e-y dy-∫ddyy∫e-y dydy+C⇒e-yx=-ye-y -e-y +C⇒e-yx+ye-y +e-y =C⇒x+y+1e-y =C⇒x+y+1=CeyHence, x+y+1=Cey is the required solution.
ix) We have,dydxcos2x=tan x-y⇒dydx+1cos2 xy=tan x sec2 x⇒dydx+y sec2 x=tan x sec2 x …..1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=sec2 xQ=tan x sec2 x∴ I.F.=e∫P dx =e∫sec2 x dx =etan xMultiplying both sides of 1 by etan x, we getetan x dydx+y sec2 x=etan xtan x sec2 x⇒etan xdydx+yetan xsec2 x=etan xtan x sec2 xIntegrating both sides with respect to x, we getetan xy=∫etan xtan x sec2 x dx+C⇒etan xy=I+C …..2Where,I=∫etan xtan x sec2 x dxPutting t= tan x, we getdt=sec2 x dx∴ I=∫tI etII dt =t∫et dt-∫ddtt∫et dtdt =t et-et =t-1et =tan x-1etan xPutting the value of I in 2, we getetan xy=tan x-1etan x+C⇒y=tan x-1+Ce-tan xHence, y=tan x-1+Ce-tan x is the required solution.
x) We have,e-ysec2y dy=dx+x dy⇒ dx=e-ysec2y dy-x dy⇒ dxdy=e-ysec2y-x⇒ dxdy+x=e-ysec2y …..1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=1Q=e-ysec2y∴ I.F.=e∫P dy =e∫dy =eyMultiplying both sides of 1 by ey, we geteydxdy+x=eye-ysec2y⇒eydxdy+eyx=sec2yIntegrating both sides with respect to y, we geteyx=∫sec2y dy+C⇒eyx=tan y+C⇒x=tan y+Ce-yHence, x=tan y+Ce-y is the required solution.
xi) We have,x log xdydx+y=2log xDividing both sides by x log x, we getdydx+yx log x=2 log xx log x⇒dydx+yx log x=2 x⇒dydx+1x log xy=2 xComparing with dydx+Py=Q, we getP=1x log xQ=2 xNow, I.F.=e∫Pdx=e∫1x log xdx =eloglog x =log xSo, the solution is given by y×I.F.=∫Q×I.F. dx+C⇒y log x=2∫1 x×log x dx+CPutting log x=t⇒1xdx=dt∴y log x=2∫t dt+C⇒y log x=2t22+C⇒y log x=t2+C⇒y log x=log x2+C ∵ log x=t⇒y=log x+Clog x
xii) We have,xdydx+2y=x2log xDividing both sides by x, we getdydx+2yx=x log xComparing with dydx+Py=Q, we getP=2xQ=x log xNow, I.F.=e∫Pdx=e∫2xdx =e2logx =x2So, the solution is given byy×I.F.=∫Q×I.F. dx+C⇒x2y=∫x3IIlog xI dx+C⇒x2y=log x∫x3 dx-∫ddxlog x∫x3 dxdx+C⇒x2y=x4log x4-∫x34dx+C⇒x2y=x4log x4-x416+C⇒y=x2log x4-x216+Cx2⇒y=x2164log x-1+Cx2
Q37.
Answer :
i) We have,y’+y=ex⇒dydx+y=ex …..1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=1 and Q=ex∴ I.F.=e∫P dx =e∫1 dx = exMultiplying both sides of 1 by I.F.=ex, we getex dydx+y=exex⇒exdydx+exy=e2xIntegrating both sides with respect to x, we gety ex=∫e2x dx+C⇒y ex=e2x2+C …..2Now, y0=12∴ 12 e0=e02+C⇒C=0Putting the value of C in 2, we getyex=e2x2⇒ex=ex2Hence, y=ex2 is the required solution.
ii) We have,xdydx-y=log x⇒dydx-yx=log xx …..1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1x and Q=log xx∴ I.F.=e∫P dx =e-∫1x dx =e-log x =1xMultiplying both sides of 1 by I.F.=1x, we get1x dydx-1xy=1x×logxx⇒1xdydx-1x2y=logxx2Integrating both sides with respect to x, we gety1x=∫1x2II×logxI dx+C⇒yx=log x∫1x2dx-∫ddxlog x∫1x2dxdx+C⇒yx=-log xx+∫1x2dx+C⇒yx=-log xx-1x+C⇒y=-log x-1+Cx …..2Now, y1=0∴ 0=-0-1+C1⇒C=1Putting the value of C in 2, we gety=-log x-1+x⇒y=x-1-log xHence, y=x-1-log x is the required solution.
iii) We have, dydx+2y=e-2xsin x …..1Clearly, it is a linear differential equation of the formdydx+Py=Qwhere P=2 and Q=e-2xsin x∴ I.F.=e∫P dx =e∫2 dx = e2xMultiplying both sides of 1 by I.F.=e2x, we gete2x dydx+2y=e2xe-2xsin x⇒e2x dydx+2y=sin xIntegrating both sides with respect to x, we getye2x=∫sin x dx+C⇒ye2x=-cos x+C …..2Now,y0=0∴ 0×e0=-cos 0+C⇒C=1Putting the value of C in 2, we getye2x=-cos x+1⇒ye2x=1-cos xHence, ye2x=1-cos x is the required solution.
iv) We have,xdydx-y=x+1e-x⇒dydx-1xy=x+1xe-x …..1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1x and Q=x+1xe-x∴ I.F.=e∫P dx =e-∫1x dx =e-log x =1xMultiplying both sides of 1 by I.F.=1x, we get1x dydx-1xy=1xx+1xe-x ⇒1x dydx-1xy=x+1x2e-xIntegrating both sides with respect to x, we get1xy=∫1x+1x2e-x dx+CPutting 1xe-x=t⇒-1xe-x-1x2e-xdx=dt⇒1x+1x2e-x dx=-dt∴1xy=∫-dt+C⇒yx=-t+C⇒yx=-e-xx+C⇒y=-e-x+Cx …..2Now, y1=0∴ 0=-e-1+C⇒C=e-1Putting the value of C in 2, we gety=-e-x+xe-1⇒y=xe-1-e-xHence, y=xe-1-e-x is the required solution.
v) We have,1+y2dx+x-e-tan-1ydy=0⇒x-e-tan-1ydydx=-1+y2⇒1+y2dxdy=-x-e-tan-1y⇒dxdy+x1+y2=e-tan-1 y1+y2 …..1Clearly, it is a linear differential equation of the form dxdy+Px=Qwhere P=11+y2 and Q=e-tan-1 y1+y2∴ I.F.=e∫P dy =e∫11+y2 dy =etan-1yMultiplying both sides of 1 by I.F.=etan-1y, we getetan-1y dxdy+x1+y2=etan-1ye-tan-1 y1+y2⇒etan-1y dxdy+x1+y2=11+y2Integrating both sides with respect to y, we getetan-1yx=∫11+y2 dy+C⇒xetan-1y=tan-1y+C …..2Now, y0=0∴ 0×e0=0+C⇒C=0Putting the value of C in 2, we getxetan-1y=tan-1y+0⇒xetan-1y=tan-1yHence, xetan-1y=tan-1y is the required solution.
vi) We have,dydx+y tan x=2x+x2tan x …..1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=tan x and Q=x2cot x+2x∴ I.F.=e∫P dx =e∫tan x dx = elogsec x=sec xMultiplying both sides of 1 by I.F.=sec x, we getsec xdydx+ytan x=sec xx2tan x+2x⇒sec xdydx+ytan x=x2tan x sec x+2x sec xIntegrating both sides with respect to x, we getysec x=∫x2tan xsec x dx+2∫xII sec xI dx +C⇒y sec x=∫x2tan x sec x dx+2sec x∫x dx-2∫ddxsec x∫x dxdx+C⇒y sec x=∫x2tan x sec x dx+x2sec x-∫x2tan x sec x dx+C⇒y sec x=x2sec x+C⇒y=x2+Ccos x …..2Now,y0=1∴ 1=0+Ccos 0⇒C=1Putting the value of C in 2, we gety=x2+cos xHence, y=x2+cos x is the required solution.
vii) We have,xdydx+y=x cos x+sin x⇒dydx+1xy=cos x+sin xx …..1Clearly, it is a linear differential equation of the formdydx+Py=Qwhere P=tan x and Q=x2cot x+2x∴ I.F.=e∫P dx =e∫1xdx = elog x=xMultiplying both sides of (1) by I.F.=x, we getxdydx+1xy=xcos x+sin xx⇒xdydx+1xy=x cos x+sin xIntegrating both sides with respect to x, we getxy=∫x cos x dx+∫sin x dx+C⇒xy=x sin x-∫1sin xdx-cos x+C⇒xy=x sin x+cos x-cos x+C⇒xy=x sin x+C …..2Now, yπ2=1∴ 1×π2=π2sinπ2+C⇒C=0Putting the value of C in 2, we getxy=x sin x⇒y=sin xHence, y=sin x is the required solution.
viii) We have, dydx+y cot x=4x cosec x …..1Clearly, it is a linear differential equation of the formdydx+Py=Qwhere P=cot x and Q=4x cosec x ∴ I.F.=e∫P dx =e∫cot x dx = elogsin x=sin xMultiplying both sides of 1 by I.F.=sin x, we getsin xdydx+y cot x=sin x4x cosec x⇒sin xdydx+y cot x=4xIntegrating both sides with respect to x, we gety sin x=4∫x dx+C⇒y sin x=2×2+C …..2Now,yπ2=0∴ 0×sinπ2=2π22+C⇒C=-π22Putting the value of C in 2, we gety sin x=2×2-π22Hence, y sin x=2×2-π22 is the required solution.
ix) We have,dydx+2y tan x=sin x …..1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=2tan x and Q=sin x∴ I.F.=e∫P dx =e2∫tan x dx = e2logsec x=sec2xMultiplying both sides of (1) by I.F.=sec2 x, we getsec2 x dydx+2y tan x=sec2 x×sin x⇒sec2 x dydx+2y tan x=tan x sec xIntegrating both sides with respect to x, we gety sec2 x=∫tan x sec x dx+C⇒y sec2 x=sec x+C …..2Now, yπ3=0∴ 0secπ32=secπ3+C⇒C=-2Putting the value of C in 2, we gety sec2 x=sec x-2⇒y=cos x-2cos2 xHence, y=cos x-2cos2 x is the required solution.
x) We have, dydx-3y cot x=sin 2x …..1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-3cot x and Q=sin 2x∴ I.F.=e∫P dx =e-3∫cot x dx = e-3logsin x=cosec3xMultiplying both sides of 1 by I.F.=cosec3x, we getcosec3xdydx-3y cot x=sin 2xcosec3x⇒cosec3xdydx-3y cot x=2cot x cosec xIntegrating both sides with respect to x, we gety cosec3x=2∫cot x cosec x dx+C⇒ ycosec3x=-2cosec x+C⇒y=-2sin2x+Csin3x …..2Now, yπ2=2∴ 2=-2sin2π2+Csin3 π2⇒C=4Putting the value of C in 2, we gety=-2sin2x+4sin3x⇒y=4sin3x-2sin2xHence, y=4sin3x-2sin2x is the required solution.
Q38.
Answer :
We have, xdydx+2y=x2 ⇒dydx+2xy=x …..1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=2x and Q=x. ∴ I.F.=e∫P dx =e∫2x dx =e2log x =x2 Multiplying both sides of 1 by I.F.=x2, we getx2dydx+2xy=x2x ⇒x2dydx+2xy=x3 Integrating both sides with respect to x, we getx2y=∫x3 dx+C⇒x2y=x44+C⇒y=x24+Cx-2Hence, y=x24+Cx-2 is the required solution.
Q39.
Answer :
We have,dydx-y=cos x …..1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1 and Q=cos x∴ I.F.=e∫P dx =e-∫ dx = e-xMultiplying both sides of 1 by I.F.=e-x, we gete-x dydx-y=e-xcos x ⇒e-xdydx-e-xy=e-xcos xIntegrating both sides with respect to x, we getye-x=∫e-xcos x dx+C⇒ye-x=I+C …..2Here,I=∫e-xcos x dx …..3⇒I=e-xsin x-∫-e-xsin x dx⇒I=e-xsin x+∫e-xsin x dx⇒I=e-xsin x-e-xcos x-∫-e-x×-cos x dx⇒I=e-xsin x-e-xcos x-∫e-xcos x dx⇒I=e-xsin x-e-xcos x-I From 3 ⇒2I=e-xsin x-cos x⇒I=e-x2sin x-cos x …..4From 2 and 4 we get⇒ye-x=e-x2sin x-cos x+C⇒y=12sin x-cos x+CexHence, y=12sin x-cos x+Cex is the required solution.
Q40.
Answer :
We have, y+3x2dxdy=x⇒dydx=y+3x2x⇒dydx-1xy=3x …..1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1x and Q=3x∴ I.F.=e∫P dx =e-∫1x dx =e-log x =1xMultiplying both sides of (1) by I.F.=1x, we get1x dydx-1xy=1x3x ⇒1xdydx-1x2y=3Integrating both sides with respect to x, we get1xy=3∫dx+C⇒yx=3x+CHence, yx=3x+C is the required solution.
Page 22.104 Ex. 22.10
Q41.
Answer :
We have,dxdy+x cot y=2y+y2cot y …..1Clearly, it is a linear differential equation of the form dxdy+Px=Qwhere P=cot y and Q=2y+y2cot y∴ I.F.=e∫P dy =e∫cot y dy = elogsin y=sin yMultiplying both sides of 1 by I.F.=sin y, we getsin ydxdy+x cot y=sin yy2cot y+2y⇒sin ydxdy+x cos y=y2cos y+2y sin yIntegrating both sides with respect to y, we getx sin y=∫y2Icos yIIdy+∫2y sin y dy+C⇒x sin y=y2∫cos ydy-∫ddyy2∫cos y dydy+∫2y sin y dy+C⇒x sin y=y2sin y-∫2y sin y dy+∫2ysin y dy+C⇒x sin y=y2sin y+CNow, y=π2 at x=0∴ 0×sin π2=π42sin π2+C⇒C=-π42Putting the value of C, we getx sin y=y2sin y-π42Hence, x sin y=y2sin y-π42 is the required solution.
Page 22.128 Ex. 22.11
Q1.
Answer :
Let r be the radius and S be the surface area of the balloon at any time t. Then,
S=4πr2⇒dSdt=8π r drdt …..1Given: dSdtα t⇒dSdt=kt, where k is any constantPutting dSdt=kt in (1), we get⇒kt=8π r drdtkt dt=8πr drIntegrating both sides, we get ∫kt dt=∫8πr dr⇒kt22=8π ×r22+C …..(2)At t=0 s, r=1 unit and at t=3 s, r=2 units Given∴ 0=8π×12+C⇒C=-4πAnd92k=8π×2+C⇒92k=12 π⇒k=83πSubstituting the values of C and k in (2), we get 8t26π=8π ×r22-4π⇒4t23=4r2-4⇒t23=r2-1⇒r2=1+t23⇒r=1+13t2
Q2.
Answer :
Let PO be the initial population and P be the population at any time t. Then,
dPdt=5P100⇒dPdt=0.05P
⇒dPP=0.05dt Integrating both sides with respect to t, we get∫dPP=∫0.05dt log P=0.05t +CNow,P=P0 at t=0 ∴ log P0=0+C⇒C=log P0Putting the value of C, we getlog P=0.05t+log P0⇒logPP0=0.05tTo find the time when the population will double, we haveP=2P0∴ log2P0P0=0.05t⇒log 2=0.05t⇒t=log 20.05=20 log 2 years
Q3.
Answer :
Let the original population be N and the population at any time t be P.
Given: dPdtαP
⇒dPdt=aP⇒dPP=adt⇒logP=at+C …..1Now, P=N at t=0Putting P=N and t=0 in 1, we getlogN=CPutting C=logN in 1, we getlogP=at+logN⇒logPN=at …..2According to the question,log2NN=25a⇒a=125log2=125×0.6931=0.0277Putting a=0.0277 in 2, we getlogPN=0.0277t …3For P=500000 and N=100000:log500000100000=0.0277t⇒t=log 50.0277=1.6090.0277=58.08 years=Approximately 58 years
Q4.
Answer :
Let the bacteria count at any time t be N.Given: dNdtα N⇒dNdt=λN⇒1NdN=λdtIntegrating both sides, we get∫1NdN=∫λdt⇒log N=λt+log C …..(1)Initially when t=0, then N=100000 Given∴log 100000=0+log C⇒log C=log 100000After 2 hours number increased by 10%Therefore, increased number=1000001+10%=110000Given: t=2, N=110000Putting t=2, N=110000 in (1), we getlog 110000=2λ+log 100000⇒12log 1110=λSubstituting the values of log C and λ in (1), we getlog N=t2log 1110+log 100000 …..(2)Now,Let t=T when N=200000Substituting these values in (2), we get log 200000=T2log 1110+log 100000⇒log 2=T2log 1110⇒T=2log 2log1110∴ The count will reach 200000 in 2log 2log1110 hours.
Q5.
Answer :
Let PO be the initial amount and P be the amount at any time t. Then,
dPdt=6P100⇒dPdt=0.06P
⇒dPP=0.06dt Integrating both sides with respect to t, we getlog P=0.06t +CNow, P=P0 at t=0 ∴ log P0=0+C⇒C=log P0Putting the value of C, we getlog P=0.06t +log P0⇒logPP0=0.06t⇒e0.06t=PP0To find the amount after 10 years, we get⇒e0.06×10=PP0⇒e0.6=PP0⇒1.822=PP0⇒P=1.822P0⇒P=1.822×1000=Rs 1822To find the time after which the amount will double, we haveP=2P0∴ log2P0P0=0.06t⇒log 2=0.06t⇒t=0.69310.06=11.55 years
Q6.
Answer :
Let the original count of bacteria be N and the count of bacteria at any time t be P.
Given: dPdtαP
⇒dPdt=aP⇒dPP=adt⇒logP=at+C …..1Now,P=N at t=0 Putting P=N and t=0 in 1, we getlogN=C Putting C=logN in 1, we getlogP=at+logN⇒logPN=at …..2According to the question,log3NN=5a⇒a=15log3=15×1.0986=0.21972Putting a=0.21972 in 2, we getlogPN=0.21972t …..3 ⇒e0.21972t=PN …..4Putting t=10 in 4 to find the bacteria after 10 hours, we get e0.21972×10=PN⇒e2.1972=PN⇒PN=9⇒P=9NTo find the time taken when the number of bacteria becomes 10 times of the number of initial population, we haveP=10N∴ log10NN=15tlog 3⇒t=5 log 10log 3
Q7.
Answer :
Let the population at any time t be P.
Given: dPdt α P
⇒dPdt=βP⇒dPP=βdt⇒logP=βt+log C …1Now, At t=1990, P=200000 and at t=2000, P=250000∴ log 200000=1990β+log C …2 log 250000=2000β+log C …3Subtracting 3 from 2, we getlog 200000-log 250000=10β⇒β=110log54Putting β=110log 54 in 2, we getlog 200000=1990×110log54+log C⇒log 200000=199log54+log C ⇒log C=log 200000-199log54 Putting β=110log 54, log C=log 200000-199 log54 and t=2010 in 1, we getlogP=110×2010log 54+log 200000-199 log54⇒logP=201 log 54+log 200000-199log54⇒logP=log 54201- log54199+log 200000⇒logP=log 5420145199+log 200000⇒logP=log 542+log 200000⇒logP=log2516×200000⇒logP=log 312500⇒P=312500
Q8.
Answer :
We have,dCdx=2+0.15x⇒dC=2+0.15xdxIntegrating both sides with respect to x, we getC=2x+0.152×2+K …..1At C0=100, we have100=20+0.15202+K⇒K=100Putting the value of T in 1, we getC=2x+0.152×2+100⇒C=0.075×2+2x+100
Q9.
Answer :
Let PO be the initial amount and P be the amount at any time t.
We have,
dPdt=8P100⇒dPdt=2P25
⇒dPP=225dtIntegrating both sides with respect to t, we getlog P=225t +C …..1Now,P=P0 at t=0 ∴ log P0=0+C⇒C=log P0Putting the value of C in 1, we getlog P=225t +log P0⇒logPP0=225t⇒e225t=PP0To find the amount after 1 year, we havee225=PP0⇒e0.08=PP0⇒1.0833=PP0⇒P=1.0833P0Percentage increase =P-P0P0×100% =1.0833P0-P0P0×100% =0.0833×100% =8.33%
Q10.
Answer :
We have,Ldidt+Ri=E⇒didt+RLi=EL …..1∴ I.F.=e∫RL dt =eRLtMultiplying both sides of (1) by I.F.=eRLt, we get eRLt didt+RLi=eRLt×EL⇒eRLtdidt+eRLtRLi=eRLt×ELIntegrating both sides with respect to t, we geteRLti=EL∫eRLt dt+C⇒eRLti=EL×LReRLt +C⇒eRLti=EReRLt +C …..2Now, i=0 at t=0∴ e0×0=ERe0 +C⇒C=-ERPutting the value of C in 2, we geteRLti=EReRLt -ER⇒i=ER -ERe-RLt⇒i=ER 1-e-RLt
Page 22.129 Ex. 22.11
Q11.
Answer :
Let the initial amount of radium be N and the amount of radium present at any time t be P.
Given: dPdtαP
⇒dPdt=-aP, where a>0⇒dPP=-adtIntegrating both sides, we get⇒logP=-at+C …..1Now, P=N at t=0Putting P=N and t=0 in 1, we getlogN=CPutting C=logN in 1, we getlogP=-at+logN⇒logNP=atAccording to the question, logNN2=at⇒log2=at⇒t=1alog2Here, a is the constant of proportionality.
Q12.
Answer :
Let the original amount of the radium be N and the amount of radium at any time t be P.
Given: dPdtαP
⇒dPdt=-aP⇒dPP=-adtIntegrating both sides, we get⇒logP=-at+C …..1Now,P=N at t=0Putting P=N and t=0 in 1, we getlogN=CPutting C=logN in 1, we getlogP=-at+logN⇒logPN=-at …..2According to the question,P=12N at t=1590logN2N=-1590a⇒-log 2=-1590a⇒a=11590log 2Putting a=11590log 2 in 2, we getlogPN=-11590log 2t PN=e-log 21590t …..3Putting t=1 in 4 to find the bacteria after 1 year, we getPN=0.9996⇒P=0.9996NPercentage of amount disapeared in 1 year =N-PN×100%=N-0.9996NN×100%=0.04%
Q13.
Answer :
According to the question,dydx=-xy⇒y dy=-x dx Integrating both sides with respect to x, we get∫y dy=-∫x dx⇒y22=-x22+CSince the curve passes through 3,-4, it satisfies the above equation.∴ -422=-322+C⇒8=-92+C⇒C=252Putting the value of C, we gety22=-x22+252⇒x2+y2=25
Q14.
Answer :
We have,
y-xdydx=y2+dydx
⇒dydxx+1=y1-y⇒dyy1-y=dxx+1
Integrating both sides, we get∫dyy1-y=∫dxx+1⇒∫1y+11-ydy=∫dxx+1⇒log y-log 1-y=log x+1+C …..1Since the curve passes throught the point 2, 2, it satisfies the equation of the curve.⇒log 2-log 1-2=log 2+1+C⇒C=log 23Putting the value of C in 1, we getlog y-log 1-y=log x+1+log 23⇒log y1-y=log 2x+13⇒y1-y=2x+13⇒y1-y=±2x+13⇒y1-y=2x+13 or y1-y=-2x+13Here, given point 2, 2 does not satisfy y1-y=2x+13But it satisfy y1-y=-2x+13∴y1-y=-2x+13⇒yy-1=2x+13⇒3y=2x+1y-1⇒3y=2xy-2x+2y-2⇒2xy-2x-y-2=0
Q15.
Answer :
The slope of the curve is given as dydx=tan θ.
Here,
θ=tan-1yx-cos2yx
∴ dydx=tantan-1yx-cos2yx⇒dydx=yx-cos2yx
Let y=vx⇒dydx=v+xdvdx∴ v+xdvdx=v-cos2v⇒xdvdx=-cos2v⇒sec2 v dv=-1xdxIntegrating both sides with respect to x, we get∫sec2 v dv=-∫1xdx⇒tan v=-log x+C⇒tan yx=-log x+CSince the curve passes through 1, π4, it satisfies the above equation.∴ tan π4=-log 1+C⇒C=1Putting the value of C, we gettan yx=-log x+1⇒tan yx=-log x+log e⇒tan yx=logex
Q16.
Answer :
Let the given curve be y = f(x). Suppose P(x,y) be a point on the curve. Equation of the tangent to the curve at P is
Y – y =dydx(X-x) , where (X, Y) is the arbitrary point on the tangent.
Putting Y=0 we get,
0 – y = dydx(X – x)Therefore, X-x=-ydxdy⇒X=x-ydxdyTherefore, cut off by the tangent on the x-axis = x-ydxdyGiven, x-ydxdy=4yTherefore, -ydxdy=4y – x⇒dxdy=x-4yy⇒dydx=yx-4y ……..(1)this is a homogeneous differential equation.
Putting y = vx and dydx=v+xdvdx in (1) we getv + xdvdx = vxx-4vxTherefore, v+xdvdx=v1-4v⇒xdvdx=v1-4v-v = 4v21-4v⇒1-4vv2dv=4dxx
Integrating on both sides we get,
∫1-4vv2dv=4∫dxxTherefore, ∫dvv2-4∫dvv=4∫dxx⇒-1v-4 log v = 4 logx + log c⇒-1v = 4 logx + log c +4 log v⇒4 log(xv) + log c = -1vputting the value of v we get4 log(x×yx) + log c = -xy⇒4 log(y) + log c = -xy⇒log (y4c) = -xy⇒y4c = e-xy
Q17.
Answer :
According to the question,dydx=y+2x⇒dydx-y=2x …..1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1 and Q=2x∴ I.F.=e∫P dx =e-∫ dx = e-xMultiplying both sides of 1 by I.F.=e-x, we gete-x dydx-y=e-x2x ⇒e-xdydx-e-xy=e-x2x Integrating both sides with respect to x, we gety e-x=2∫e-xIIxI dx+C⇒ye-x=2x∫e-xdx-2∫ddxx∫e-xdxdx+C⇒ye-x=-2xe-x-2e-x+C …..2Since the curve passes through origin, we have0×e0=-2×0×e0-2e0+C⇒C=2Putting the value of C in 2, we getye-x=-2xe-x-2e-x+2⇒y=-2x-2+2ex⇒y+2x+1=2exDISCLAIMER: In the question it should be ex instead of e2x.
Q18.
Answer :
The slope of the curve is given as dydx=tan θ.
Here,
θ=tan-1 2x+3y∴ dydx=tantan-1 2x+3y⇒dydx=2x+3y
⇒dydx-3y=2x …..1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-3 and Q=2x∴ I.F.=e∫P dx =e∫-3 dx = e-3xMultiplying both sides of (1), by I.F.=e-3x, we gete-3x dydx-3y=e-3x.2x⇒e-3x dydx-3y=2xe-3xIntegrating both sides with respect to x, we gety e-3x=2∫xe-3x dx+C⇒y e-3x=2∫xIe-3xII dx+C⇒y e-3x=2x∫e-3x dx-2∫ddxx∫e-3x dxdx+C⇒y e-3x=-2xe-3×3+2×13∫e-3x dx+C⇒y e-3x=-23xe-3x-2×19e-3x+C⇒y e-3x=-23xe-3x-29e-3x+CSince the curve passes through 1, 2, it satisfies the above equation.∴ 2e-3=-23e-3-29e-3+C⇒C=2e-3+23e-3+29e-3⇒C=269e-3Putting the value of C, we gety e-3x=-23x-29e-3x+269e-3
Q19.
Answer :
Portion of the x-axis cut off between the origin and tangent at a point =x−y dxdy=OT
It is given, OT = 2x
∴ x−y dxdy=2x−x=ydxdy-∫dxx=∫dyy∴ xy=k
Since the curve passes through the point (1, 2)
⇒ at x = 1 ⇒ y = 2
∴ k = 2
∴ xy = 2
Q20.
Answer :
We have,xx+1dydx-y=xx+1⇒dydx-yxx+1=1Comparing with dydx+Py=Q, we getP=-1xx+1Q=1Now,I.F.=e-∫1xx+1dx =e-∫1x-1x+1dx =e-logxx+1 =x+1x So, the solution is given byy×I.F.=∫Q×I.F. dx +C⇒x+1xy=∫x+1x dx +C⇒x+1xy=∫dx+∫1xdx +C⇒x+1xy=x+log x+CSince the curve passes throught the point 1, 0, it satisfies the equation of the curve.⇒1+110=1+log 1+C⇒C=-1Putting the value of C in the equation of the curve, we getx+1xy=x+log x-1⇒y=xx+1x+log x-1
Q21.
Answer :
According to the question,
dydx=2yx
⇒12ydy=1xdxIntegrating both sides with respect to x, we get∫12ydy=∫1xdx⇒12log y=log x+CSince the curve passes through 3,-4, it satisfies the above equation.∴ 1 2log -4=log 3+C⇒log 2-log 3=C⇒C=log 23Putting the value of C, we getlog y=2log x+2log 23⇒log y=log 49×2⇒y=±49×2⇒9y-4×2=0 or 9y+4×2=0 The given point does not satisfy the equation 9y-4×2=0. ∴ 9y+4×2=0
Q22.
Answer :
According to the question,
dydx=x+3y-1
⇒dydx-3y=x-1
Comparing with dydx+Py=Q, we getP=-3 Q=x-1Now, I.F.=e-∫3dx =e-3xSo, the solution is given byy×I.F.=∫Q×I.F. dx +C⇒ye-3x=∫x-1e-3x dx +C⇒ye-3x =∫xIe-3xIIdx-∫e-3x dx+C⇒ye-3x=x∫e-3x dx-∫ddxx∫e-3x dxdx-∫e-3x dx+C⇒ye-3x=-13xe-3x+13∫e-3x dx-∫e-3x dx+C⇒ye-3x=-13xe-3x-19e-3x +13e-3x+C⇒y=-13x-19 +13+Ce3x⇒y=-13x +29+Ce3xSince the curve passes throught the origin, it satisfies the equation of the curve.⇒0=-0+29+Ce0⇒C=-29Putting the value of C in the equation of the curve, we gety=-13x +291-e3x⇒y+13x=291-e3x⇒33y+x=21-e3x
Q23.
Answer :
According to the question,
dydx=x+xy⇒dydx=x1+y
⇒11+ydy=x dxIntegrating both sides with respect to x, we get∫11+ydy=∫x dx⇒log 1+y=x22+CSince the curve passes through 0, 1, it satisfies the above equation.∴ log 1+1=02+C⇒C=log 2Putting the value of C, we getlog 1+y=x22+log 2⇒log 1+y2=x22⇒1+y2=ex22⇒y+1=2ex22
Q24.
Answer :
Tangent at P(x, y) is given by Y – y = dydx(X-x)
If p be the perpendicular from the origin, then
p = xdydx-y1+dydx2=x (given)⇒x2dydx2-2xydydx + y2=x2 +x2 dydx2⇒y2-2xydydx-x2 =0 Hence proved.Now, y2-2xydydx-x2 =0 ⇒dydx = y2-x22xy⇒2xydydx-y2 =-x2 ⇒2ydydx-y2 x=-x Let y2=v⇒dvdx-vx=-x
Multiplying by the integrating factor e∫-1xdx=1x
v.1x=∫-x.1xdx + c= -x+c⇒y2x2=-x+c⇒x2 + y2=cx
Q25.
Answer :
It is given that the distance between the foot of ordinate of point of contact (A) and point of intersection of tangent with x-axis (T) = 2x
Coordinate of T=x−ydxdy, 0AT=x−x−y dxdy=2xEquation of tangent, Y−y=dydx(X−x) ⇒y-0=dydxx−x−y dxdy⇒y dxdy=2x⇒∫dxx=2∫dyy⇒lnx=lny2+lncx=cy2As the circle passes through 1, 2.1=c×22⇒c=14⇒4x=y2
Q26.
Answer :
Let P (x, y) be any point on the curve. The equation of the normal at P (x, y) to the given curve is given as
Y-y=-1dydxX-x
It is given that the curve passes through the point (3, 0). Then,
0-y=-1dydx3-x⇒-y=-1dydx3-x⇒ydydx=3-x⇒y dy=3-xdx⇒y22=3x-x22+C …..1Since the curve passes through the point 3, 4, it satisfies the equation.⇒422=33-322+C⇒C=8-9+92⇒C=92-1=72Putting the value of C in 1, we gety22=3x-x22+72⇒y2=6x-x2+7⇒x2+y2-6x-7=0
Q27.
Answer :
Let the original count of bacteria be N and the count of bacteria at any time t be P.
Given: dPdtαP
⇒dPdt=aP⇒dPP=adt⇒log P=at+C …..1Now, P=N at t=0Putting P=N and t=0 in 1, we getlog N=CPutting C=log N in 1, we getlog P=at+log N⇒log PN=at …..2According to the question,log 2NN=6a⇒a=16log 2Putting a=16log 2 in 2, we getlog PN=t6log 2 …..3Putting t=18 in 3 to find the bacteria after 18 hours, we getlog PN=186 log 2⇒log PN=3log 2⇒log PN=log 8⇒PN=8⇒P=8N
Page 22.130 Ex. 22.11
Q28.
Answer :
Let the original amount of radium be N and the amount of radium at any time t be P.
Given: dPdtαP
⇒dPdt=-aP⇒dPP=-a dtIntegrating both sides, we get⇒log P=-at+C …..1Now,P=N when t=0 Putting P=N and t=0 in 1, we getlog N=CPutting C=log N in 1, we getlog P=-at+log N⇒log PN=-at …..2According to the question,P=98.9100N=0.989N at t=25∴log 0.989NN=-25a⇒a=-125log 0.989Putting a=-125log 0.989 in 2, we getlogPN=125log 0.989tTo find the time when the radium becomes half of its quantity, we haveN=12P∴log NN2=125log 0.989t⇒log 2=125log 0.989t ⇒t=25log 2log 0.989=25×0.69310.01106=1566.68≈1567 approx.
Q29.
Answer :
We have,
dydx=x2+y22xyLet y=vxdydx=v+xdvdx∴ v+xdvdx=x2+v2x22vx2⇒xdvdx=1+v22v-v⇒x dvdx=1+v2-2v22v⇒xdvdx=1-v22v⇒2v1-v2dv=1xdx
Integrating both sides, we get∫2v1-v2dy=∫1xdx⇒-log 1-v2=log x-log C⇒log 1-v2C=-log x⇒1-v2=Cx⇒1-yx2=Cx⇒x2-y2x2=Cx⇒x2-y2=Cx
Thus, x2-y2=Cx is the equation of the rectangular hyperbola.
Q30.
Answer :
According to the question,
dydx=x+y
⇒dydx-y=x
Comparing with dydx+Py=Q, we getP=-1Q=xNow, I.F.=e-∫dx =e-xSo, the solution is given byy×I.F.=∫Q×I.F. dx +C⇒ye-x=∫xIe-xIIdx+C⇒ye-x=x∫e-x dx-∫ddxx∫e-x dxdx+C⇒ye-x=-xe-x+∫e-x dx+C⇒ye-x=-xe-x-e-x+CSince the curve passes throught the origin, it satisfies the equation of the curve.⇒0e0=-0e0-e0+C⇒C=1Putting the value of C in the equation of the curve, we getye-x=-xe-x-e-x+1⇒ye-x+xe-x+e-x=1⇒y+x+1e-x=1⇒x+y+1=ex
Q31.
Answer :
According to the question,
dydx=x+xy
⇒dydx-xy=x
Comparing with dydx+Py=Q, we getP=-xQ=xNow, I.F.=e-∫xdx =e-x22So, the solution is given byy×I.F.=∫Q×I.F. dx +C⇒ye-x22=∫xe-x22dx+C⇒ye-x22=I+CNow,I=∫xe-x22dxPutting -x22=t, we get-xdx=dt∴ I=-∫etdt⇒I=-et⇒I=-e-x22∴ ye-x22=-e-x22+C Since the curve passes throught the point 0, 1, it satisfies the equation of the curve.⇒1e0=-e0+C⇒C=2Putting the value of C in the equation of the curve, we getye-x22=-e-x22+2⇒y=-1+2ex22
Q32.
Answer :
According to the question,
dydx=x2
⇒dy=x2dxIntegrating both sides with respect to x, we get∫dy=∫x2dx⇒y=x33+CSince the curve passes through -1, 1, it satisfies the above equation.∴ 1=-13+C⇒C=1+13⇒C=43Putting the value of C, we gety=x33+43⇒3y=x3+4
Q33.
Answer :
According to the question,
ydydx=x⇒y dy=x dxIntegrating both sides with respect to x, we get∫y dy=∫x dx⇒y22=x22+CSince the curve passes through 0, a, it satisfies the above equation.∴ a22=02+C⇒C=a22Putting the value of C, we gety22=x22+a22⇒x2-y2=-a2
Q34.
Answer :
Let P(x, y) be any point on the curve. Then slope of the tangent at P is dydx.
It is given that the slope of the tangent at P(x,y) is equal to the ordinate i.e y.
Therefore dydx = y
⇒1ydy = dx⇒log y = x + log C⇒log y = log ex + log C⇒y = Cex
Since, the curve passes through (1,1). Therefore, x=1 and y=1 .
Putting these values in equation obtained above we get,
1=Ce1⇒C=1eputting these values in the equation we get,y=ex-1
Page 22.131 (Very Short Answers)
Q1.
Answer :
Differential equation:
An equation containing an independent variable, a dependent variable and differential coefficients of the dependent variable with respect to the independent variable is called a differential equation.
for example: dydx=ex+y
Q2.
Answer :
Order of differential equation:
The order of a differential equation is the order of its highest order derivative that apears in the equation.
example: d2ydx2-4dydx=2y
order of the differential equation is 2.
Page 22.132 (Very Short Answers)
Q3.
Answer :
Degree of differential equation:
The degree of a differential equation is the power of the highest order derivative occurring in a differential equation when it is written as a polynomial in differential coefficients.
example: d2ydx22-4dydx=2y
the degree of the given differential equation is 2
Q4.
Answer :
We have,y= Cx+5 …..1⇒dydx=CSubstituting the value of C in 1, we gety= dydx×x+5⇒xdydx-y+5=0 Hence, xdydx-y+5=0 is the differential equation representing the family of straight lines y=Cx+5, where C is an arbitary constant.
Q5.
Answer :
We have,x2-y2=C2Differentiating with respect to x, we get2x-2ydydx=0⇒2x=2ydydx⇒x dx=y dy⇒x dx-y dy=0Hence, x dx-y dy=0 is the required differential equation.
Q6.
Answer :
We have,xy=C2Differentiating with respect to x, we getxdydx+y=0⇒xdydx=-y⇒x dy=-y dx⇒x dy+y dx=0Hence, x dy+y dx=0 is the required differential equation.
Q7.
Answer :
We have,a2d2ydx2=1+dydx21/4a2d2ydx24=1+dydx2Degree of the differential equation is the degree of the highest order derivative.Therefore, the degree must be 4.
Q8.
Answer :
1+dydx2=7 d2ydx23The order of a differential equation is the order of its highest order derivatives.Here, the required order is 2.
Q9.
Answer :
We have,y=xdydx+a1+dydx2y-xdydx=a1+dydx2Squaring both sides, we gety-xdydx2=a1+dydx22y2+x2dydx2-2xydydx=a21+dydx2y2+dydx2x2-a2-2xydydx=a2From the above equation, we see that the highest order is 1. So, its order is 1 and the power of the highest order derivative is 2.Thus, it is a differential equation of order 1 and degree 2.
Q10.
Answer :
We have,d2ydx2+xdydx2=2×2 log d2ydx2⇒d2ydx2+xdydx2-2×2 log d2ydx2=0Here, we observe that LHS of the differential equation cannot be expressed as a polynomial in dydx. So, its degree is not defined.
Q11.
Answer :
The equation of the family of circles touching x-axis at the origin is
x-02+y-a2=a2x2+y2-2ay=0 …..1Here, a is the parameter.Since this equation contains only one arbitary constant, we differentiate it only once. Differentiating with respect to x, we get2x+2ydydx-2adydx=0a=x+ydydxdydx …..2Putting the value of a from (2) in (1), we get x2+y2=2yx+ydydxdydxx2-y2dydx=2xySo, this is the required differential equation.Here, order of the differential equation is 1.
Q12.
Answer :
The equation of the non-horizontal lines in a plane isy=mx+c, where m is the slope and c is the intercept on y-axis.Differentiating with respect to x, we getdydx=m⇒d2ydx2=0This is the required differential equation. Here, we observe that the order of the required differential equation is 2.
Q13.
Answer :
It is given that sin x is the integrating factor of the differential equation dydx+Py=Q.∴ e∫Pdx=sin x⇒∫P dx=log sin x⇒∫P dx=∫cot x dx ∵∫cot x dx=log sin x+C⇒P=cot x
Q15.
Answer :
y=a cos x+b sinx +c e-xHere, we see that there are three arbitary constants. Therefore, we differentiate it three times to get rid of all three arbitrary constants.Hence, the order of the differential equation is 3.
Q16.
Answer :
y=C1+C2ex+C3e-2x+C4the given equation can be reduced to:y=C1+C2ex+C3(e-2x×ec4)Here, ec4 will be a constant.We have 3 constants as C1, C2 and C3.and a differential equation of order n always contains exactly n essential arbitrary constants.Hence, the order of the required differntial equation is 3.
Q17.
Answer :
5xdydx2-d2ydx2-6y=log x
Here, we see that the highest order derivative is d2ydx2 and its power is 1.
Therefore, the given differential equation is of first degree.
Q18.
Answer :
dydx4+3xd2ydx2=0
The highest order derivative is d2ydx2 and its power is 1.
Therefore, the given differential equation is of first degree.
Q19.
Answer :
d2ydx23+ydydx4+x3=0
The highest order derivative is d2ydx2 and its power is 3.
Therefore, the degree of given differential equation is 3.
Q20.
Answer :
We have,y=mx …..1 Differentiating with respect to x⇒dydx=mSubstituting the value of m=dydx in eq 1 we get ,y=xdydxHence, y=xdydx is the required differential equation .
Q21.
Answer :
x3d2ydx22+xdydx4=0
Here, the highest order derivative is d2ydx2 and its power is 2.
Therefore, degree of given differential equation is 2.
Page 22.133 (Multiple Choice Questions)
Q1.
Answer :
(c) log x
We have,
(x log x) dydx+y=2 log x
Dividing both sides by x log x, we get
dydx+yxlogx=2 log xxlogx⇒dydx+yxlogx=2 x⇒dydx+1xlogxy=2 xComparing with dydx+Py=Q, we getP=1xlogxQ=2 xNow, I.F.=e∫Pdx=e∫1x log xdx =eloglog x =log x
Q2.
Answer :
(b) y = kx
We have,
dydx=y x⇒1ydy=1 xdxIntegrating both sides, we get∫1ydy=∫1 xdx⇒log y=log x+log k⇒log y-log x=log k⇒logyx=log k⇒yx= k⇒y= kx
Q3.
Answer :
(b) sec x
We have,
cos xdydx+y sin x=1
Dividing both sides by cos x, we get
dydx+sin xcos xy=1cos x⇒dydx+tan xy=1cos xComparing with dydx+Py=Q, we getP=tan xQ=1 cos xNow, I.F.=e∫tan xdx=elogsec x =sec x
Q4.
Answer :
(b) 2
We have,
d2ydx22-dydx=y3
The highest order derivative is d2yd2x and its power is 2.Hence, the degree is 2.
Q5.
Answer :
We have,
5+dydx253=x5 d2yd2xTaking Cube power on both sides, we get5+dydx25=x15 d2yd2x3The highest order derivative is d2yd2x and its power is 3.Hence, the degree is 3.
Disclaimer: The correct option is not given in the question.
Q6.
Answer :
(d) y sin x = x + C
We have,dydx+y cot x=cosec xdydx+ycot x=cosec xComparing with dydx+Py=Q, we getP=cot x Q=cosec xNow, I.F.=e∫cot x dx=elogsin x =sin xSo, the solution is given byysinx=∫sin x×cosec x dx+C⇒y sin x=x +C
Q7.
Answer :
(c) y” = −ω2 y
We have,
y = A cos ωt + B sin ωt …..(1)
Differentiating both sides of (1) with respect to x, we get
dydt=-Aω sin ωt+Bω cos ωt …..(2)
Differentiating both sides of (2) again with respect to x, we get
d2ydt2=-Aω2 cos ωt-Bω2 sin ωt⇒d2ydt2=-ω2A cos ωt+B sin ωt⇒d2ydt2=-ω2y Using 1∴y”=-ω2y
Q8.
Answer :
(a) x2 = y
We have,
dydx=2yx⇒12×1ydy=1 xdxIntegrating both sides, we get12∫1ydy=∫1 xdx⇒12log y=log x+log C⇒log y12-log x=log C⇒logyx=log C⇒yx=C⇒y=Cx …..1As 1 passes through (1, 1), we get∴1=CPutting the value of C in 1, we gety=x⇒y=x2
Q9.
Answer :
(c) 5
The given equation can be reduced to : y=c1cos(2x+c2)-(c)ax×ac5+c6sin(x-c7)where c=c3+c4 and ac5 will be a constant
There are 5 constants(c1, c2, c, c6, c7) in the given differential equation.
Hence, the order of the differential equation is 5.
Q10.
Answer :
(b) a = −b
We have,
dydx=ax+gby+f⇒by+fdy=ax+gdxIntegrating both sides, we get∫by+fdy=∫ax+gdx⇒by22+fy=ax22+gx+C⇒by22+fy-ax22-gx=C⇒by2+2fy-ax2-2gx-2C=0The above equation represents a circle.Therefore, the coffecients of x2 and y2 must be equal. i.e. -a=b⇒a=-b
Q11.
Answer :
(a) y=1×2
We have,
dydx+2y x=0⇒dydx=-2y x⇒12×1ydy=-1 xdxIntegrating both sides, we get12∫1ydy=-∫1 xdx⇒12log y=-log x+log C⇒log y12+log x=log C⇒logyx=log C⇒yx=C …..1As1 satisfies y1=1, we get1=CPutting the value of C in 1, we getyx=1⇒y=1×2
Q12.
Answer :
(a) y = xex+c
We have,
dydx-yx+1x=0⇒dydx=yx+1x⇒dyy=x+1xdxIntegrating both sides, we get∫dyy=∫x+1xdx⇒∫dyy=∫dx+∫1xdx⇒log y=x+log x+C⇒log y-log x=x+C⇒log yx=x+C⇒yx=ex+C⇒y=xex+C
Page 22.134 (Multiple Choice Questions)
Q13.
Answer :
(a) 1
The order of a differential equation depends on the number of arbitrary constants in it.
Since 1-x4+1-y4=ax2-y2 contains only 1 constant, the order of the differential equation is 1.
Q14.
Answer :
(b) y = C1 eC2x + C3
y1y3=y22y3y2=y2y1⇒d3ydx3d2ydx2=d2ydx2dydx⇒∫ddxd2ydx2d2ydx2=∫ddxdydxdydx⇒lnd2ydx2=lndydx+ln C4⇒d2ydx2=C4dydx⇒∫ddxdydxdydx=∫C4 dxlndydx=C4x+C5⇒dydx=eC4x+C5∫dy=∫ eC4x+C5 dxy=eC4x+C5C4+C6y=eC4x.eC5C4+C6⇒y=C1eC2x+C3where,C1=eC5C4C4=C2C6=C3
Q15.
Answer :
(b) g (x) + log {1 + y − g (x)} = C
We have,dydx+y g’x=gxg’x …..1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=g’x and Q=gxg’x. ∴ I.F.=e∫P dx =e∫g’x dx = egxMultiplying both sides of (1) by I.F., we get egx dydx+yg’x= egx gxg’x⇒ egxdydx+ egxy g’x= egxgxg’xIntegrating both sides with respect to x, we gety egx=∫egxgxg’x dx+K⇒y egx=I+K where I=∫egxgxg’x dxNow, I=∫egxgxg’x dxPutting gx=t, we getg’x dx=dt∴ I=∫tIetII dt =t∫et dt-∫ddxt∫et dtdt =tet-et =gxegx-egx∴ y egx=gxegx-egx+K⇒y egx+egx-gxegx=K⇒y+1-gx=Ke-gxTaking log on both sides, we getlogy+1-gx=-gx+log K⇒gx+log1+y-gx=C Where, C=log K
Q16.
Answer :
d y=tanx+x22
We have,
dydx=1+x+y2+xy2⇒dydx=x+1+y2x+1⇒dydx=x+11+y2⇒dy1+y2=x+1dxIntegrating both sides, we get∫dy1+y2=∫x+1dx⇒tan-1 y=x22+x+C …..1Now, y0=0∴ tan-1 0=02+0+C⇒C=0Putting the value of C in 1, we gettan-1 y=x22+x⇒y=tanx22+x
Q17.
Answer :
(a) y”y’+y’y-1x=0
We have,
x2a2+y2b2=C …..1
Differentiating with respect to x, we get
2xa2+2yb2y’=0⇒xa2+yb2y’=0 …..2Again differentiating with respect to x, we get⇒1a2+1b2y’2+yb2y”=0 …..3Multiplying throughout by x, we getxa2+xb2y’2+xyb2y”=0 …..4Subtracting 2 from 4, we get1b2xy’2+xyy”-yy’=0 ⇒xy’2+xyy”-yy’=0Dividing both sides by xyy’, we gety’y+y”y’-1x=0⇒y”y’+y’y-1x=0
Q18.
Answer :
(a) x (y + cos x) = sin x + C
We have,dydx+yx=sin x⇒dydx+1xy=sin x …..1Comparing with dydx+Py=Q, we getP=1x Q=sin xNow,I.F.=e∫1xdx =elogx =xTherefore, integration of 1 is given byy×I.F.=∫x2×I.F. dx+C ⇒yx=∫xI sin xIIdx+C⇒yx=x∫sin x dx-∫ddxx∫sin x dxdx+C⇒yx=-x cos x+∫cos x dx+C⇒yx+x cos x=sin x+C⇒xy+cos x=sin x+C
Q19.
Answer :
(c) y3 + 2x −3x2 y = 0
We have,
yx+y3dx=xy3-xdyHere, xy+y4dx=xy3-x2dy⇒xydx+y4dx-xy3dy+x2dy=0⇒xydx+xdy+y3ydx-xdy=0⇒xdxy+x2y3ydx-xdyx2=0 ⇒xdxy-x2y3xdy-ydxx2=0 ⇒dxyx2y2-yxdyx=0 ∵Dividing the whole equation by x3y2⇒dxyx2y2=yxdyx
Integrating both sides we get,
⇒∫dxyx2y2=∫yxdyx⇒-1xy=yx22-c∴-1xy-12yx2-c=0∴1xy+12y2x2+c=0∴y3+2x+2cx2y=0
It is given that the curves passes through (1, 1).
Hence,
y3+2x+2cx2y=013+21+2c11=01+2+2c=02c=-3c=-32
∴ The required curve is y3+2x-2×32x2y=0
∴y3+2x-3x2y=0
Q20.
Answer :
(d) parabolas
We have,
2xdydx-y=3⇒2xdydx=3+y⇒13+ydy=12xdxIntegrating both sides, we get∫13+ydy=12∫1xdx⇒log 3+y=12log x+log C⇒log 3+y-log x12=log C⇒log 3+yx=log C⇒3+yx=C⇒3+y=CxSquaring both sides, we get3+y2=Cx …..1Thus, 1 represents the equation of parabolas.
Q21.
Answer :
(b) sinyx=Cx
We have,
xdydx=y+x tanyx⇒dydx=yx+tanyx …..1Let y=vx⇒dydx=v+xdvdxPutting the above value in 1, we getv+xdvdx=v+tan v⇒xdvdx=tan v⇒dvtan v=dxxIntegrating both sides, we getlog sin v= log x+log C⇒log sin v- log x=log C⇒logsin vx=log C⇒sin vx=C⇒sin v=Cx⇒sinyx=Cx ∵ y=vx
Q22.
Answer :
(b) xyy2 + xy12 − yy1 = 0
We have,
ax2 + by2 = 1 …..(1)
Differentiating both sides of (1) with respect to x, we get
2ax+2bydydx=0 …..2
Differentiating both sides of (2) with respect to x, we get
2a+2bdydx2+2byd2ydx2=0⇒2byd2ydx2+dydx2=-2a⇒yd2ydx2+dydx2=-2a2b⇒yd2ydx2+dydx2=–yxdydx Using 2⇒xyd2ydx2+dydx2=ydydx⇒xyd2ydx2+xdydx2=ydydx⇒xyd2ydx2+xdydx2-ydydx=0⇒xyy2+xy12-yy1=0
Page 22.135 (Multiple Choice Questions)
Q23.
Answer :
(d) y ln y = xy1
We have,
y = eCx
Taking ln on both sides, we get
ln y = Cx ln e
⇒ln y=Cx …..1
Differentiating both sides of (1) with respect to x, we get
1yy1=C
Substituting the value of C in (1), we get
ln y=y1yx⇒y ln y=y1x
Q24.
Answer :
(c) u = (log z)-1
Given dzdx+zxlog z=zx2log z2 …..1Let u=log z-1dudx=-1log z2×1z×dzdxdzdx=-zlog z2 dudxSubstituting the value of dzdx from equation (1) we get,∴-z log z2 dudx+zxlog z=zx2log z2dudx-1x1log z=-1x2dudx-1xlog z-1=-1x2dudx-1xu=-1x2It can be written as,dudx+pxu=Qxwhere, px=-1x qx=-1×2
The correct option is C.
Q25.
Answer :
a ϕyx=kx
We have,dydx=yx+ϕyxϕ’yxLet y=vx⇒dydx=v+xdvdx∴ v+xdvdx=v+ϕvϕ’v⇒xdvdx=ϕvϕ’v⇒ϕvϕ’vdv=1xdxIntegrating both sides, we get∫ϕ’vϕvdv=∫1xdx⇒log ϕv=log x+log k⇒log ϕyx-log x=log k⇒log ϕyxx=log k⇒ ϕyxx= k⇒ϕyx=kx
Q26.
Answer :
(b) m = 3, n = 2
We have,y25+4y23y3+y3=x2-1⇒y3y25+4y23+y32=y3x2-1The highest order derivative is y3 and its highest exponent in this equation is 2.Therefore, order is 3 and degree is 2.Hence, m=3, n=2
Q27.
Answer :
(d) (x − C) ex+y + 1 = 0
We have,dydx+1=ex+yLet x+y=v⇒1+dydx=dvdx⇒dydx+1=dvdx∴ dvdx=ev⇒e-vdv=dxIntegrating both sides, we get-e-v=x-C⇒-1=evx-C⇒x-Cex+y+1=0
Q28.
Answer :
(d) x3 + y3 = 12x + C
We have,x2+y2dydx=4⇒y2dydx=4-x2⇒y2dy=4-x2dxIntegrating both sides, we get∫y2dy=∫4-x2dx⇒y33=4x-x33+D⇒y3=12x-x3+3D⇒x3+y3=12x+C, where C=3D
Q29.
Answer :
(a) x = Cy2
Subtangent=ydydx
It is given that subtangent at any point of a curve is double of the abscissa.
∴ydydx=2xy=2xdydx∫dxx=2∫dyylnx=2lny+alnx=lny2+lnclnx=lncy2x=cy2
Q30.
Answer :
(a) x2 − 1 = C (1 + y2)
We have,
x dx + y dy = x2y dy − y2x dx
⇒x+xy2dx=x2y-ydy⇒xx2-1dx=y1+y2dy⇒2x2x2-1dx=2y21+y2dyIntegrating both sides, we get12∫2y1+y2dy=12∫2xx2-1dx⇒12log1+y2=12logx2-1-12logC⇒log1+y2=logx2-1-logC⇒log1+y2=logx2-1C⇒1+y2=x2-1C⇒C1+y2=x2-1
Q31.
Answer :
d y=1-x1+x
We have,x2+1dydx+(y2+1)=0⇒x2+1dydx=-y2+1⇒1y2+1dy=-1 x2+1dxIntegrating both sides, we get∫1y2+1dy=-∫1 x2+1dx⇒tan-1 y=-tan-1 x+tan-1 C⇒tan-1 y+tan-1 x=tan-1 C⇒tan-1x+y1-xy=tan-1 C⇒x+y1-xy=CDisclaimer: The initial value conditions are not given, so the final answer will be obatined only ifC=1. So, ⇒x+y=1-xy⇒y+xy=1-x⇒y1+x=1-x⇒y=1-x1+x
Q32.
Answer :
(b) 2y − x3 = cx
We have,
xdydx-y=x2
⇒dydx-1xy=x2Comparing with dydx+Py=Q, we getP=-1x Q=x2Now,I.F.=e-∫1xdx =e-logx =elog1x =1xy×I.F=∫x2×I.Fdx+C ⇒y1x=∫x2×1xdx+C⇒y1x=∫xdx+C⇒y1x=x22+C⇒2y-x3=Cx
Q33.
Answer :
(c) k < 0
We have,⇒dydx-ky=0⇒dydx=ky⇒1ydy=k dxIntegrating both sides, we get∫1ydy=k∫dx⇒logy=kx+C …..1Now, y0=1∴ C=0Putting C=0 in 1, we getlogy=kx⇒ekx=yAccording to the question,ek∞=0Since e-∞=0∴ k<0.
Page 22.136 (Multiple Choice Questions)
Q34.
Answer :
(d) tan-1y + tan-1x = tan−1C
We have,
1+x2dydx+1+y2=0⇒1+x2dydx=-1+y2⇒11+y2dy=-1 1+x2dxIntegrating both sides we get,∫11+y2dy=-∫1 1+x2dx⇒tan-1y=-tan-1x+tan-1C⇒tan-1y+tan-1x=tan-1C
Q35.
Answer :
b) tan-1yx=log x+C
We have,
dydx=x2+xy+y2x2 1
This is homogenous differential equation.
Let y=vx⇒dydx=v+xdvdxNow, putting dydx=v+xdvdx and y=vx in 1, we getv+xdvdx=x2+x2v+x2v2x2⇒v+xdvdx=1+v+v2⇒xdvdx=1+v2⇒11+v2dv=1xdxIntegrating both sides we get,∫11+v2dv=∫1xdx⇒tan-1 v=log x+C⇒tan-1 yx=log x+C
Q36.
Answer :
(d) z = y1-n
We have,
dydx+Py=Qyn⇒y-ndydx+Py1-n=Q …..1Put z=y1-nIntegrating both sides with respect to x, we getdzdx=1-ny-ndydx⇒y-ndydx=11-ndzdxNow, 1 becomes11-ndzdx+Pz=Q⇒dzdx+P1-nz=Q1-nWhich is linear form of differential equation.Therefore, the given differential equation can be reduce to linear form by the substitution,z=y1-n
Q37.
Answer :
(c) p > q
We have,
ydydx+x3d2ydx2+xy=cos x
The highest order derivative is d2yd2x and it’s degree is 1So, the order is 2 and the degree is 1.∴p=2 and q=1Clearly, p>q
Q38.
Answer :
(c) log x
We have,
x log xdydx+y=2 log x
Dividing both sides by (x log x) we get,
dydx+yx logx=2 log xx logx⇒dydx+yx logx=2 x⇒dydx+1x logxy=2 xComparing with dydx+Py=Q we get,P=1x logx and Q=2 xNow, I.F=e∫Pdx=e∫1xlogxdx =eloglog x =log x
Q39.
Answer :
(a) sec x + tan x
We have,
dydx+y sec x=tan xComparing with dydx+Py=Q, we getP=sec x Q=tan xNow,I.F.=e∫sec xdx=elogsec x+tan x=sec x+tan x
Q40.
Answer :
(c) sec x
We have,
cos xdydx+y sin x=1
Dividing both sides by cos x, we get
dydx+sin xcos xy=1cos x⇒dydx+tan xy=1 cos xComparing with dydx+Py=Q, we getP=tan xQ=2 cos xNow,I.F.=e∫tan xdx=elogsec x=sec x
Q41.
Answer :
(d) not defined
We have,
d2ydx23+dydx2+sindydx+1=0
The highest order derivative in this equation is d2yd2x.But the equation cannot be expressed as a polynomial in differential coefficient.Hence, the degree is not defined.
Q42.
Answer :
(a) 2
We have,
2x2d2ydx2-3dydx+y=0
Here, the highest order derivative is d2yd2x.Hence, the order is 2.
Q43.
Answer :
(d) 4
The number of arbitrary constants in the general solution of a differential equation of order n is n.
Thus, the number of arbitrary constants in the general solution of differential equation of fourth order is 4.
Q44.
Answer :
(d) 0
The number of arbitrary constants in the particular solution of a differential equation is always zero.
Q45.
Answer :
b) d2ydx2-y=0
We have,
y=C1ex+C2e-x …..1
Differentiating both sides of (1) with respect to x, we get
dydx=C1ex-C2e-x …..2
Differentiating both sides of (2) with respect to x, we get
d2ydx2=C1ex+C2e-x⇒d2ydx2=y Using 1 and 2⇒d2ydx2-y =0
Page 22.137 (Multiple Choice Questions)
Q46.
Answer :
c) d2ydx2-x2dydx+xy=0
We have,
y = x …..(1)
Differentiating both sides of (1) with respect to x, we get
dydx=1 …..2Differentiating again with respect to x, we get⇒d2ydx2=0⇒d2ydx2+x2=x2⇒d2ydx2+x×x=x2×1⇒d2ydx2+xy=x2×1 Using 1⇒d2ydx2+xy=x2dydx Using 2⇒d2ydx2-x2dydx+xy=0
Q47.
Answer :
(a) ex + e-y = C
We have,
dydx=ex+y⇒dydx=ex×ey⇒e-ydy=exdxIntegrating both sides, we get∫e-ydy=∫exdx⇒-e-y=ex+D⇒ex+e-y=-D⇒ex+e-y=C Where, C=-D
Q48.
Answer :
(c) x = vy
A homogeneous differential equation of the form dxdy=hxy can be solved by substituting x = vy.
Q49.
Answer :
(d) y2 dx + (x2 − xy − y2) dy = 0
A differential equation is said to be homogenous if all the terms in the equation have equal degree and it can be written in the form dydx=fx, ygx, y.
In (a), (b) and (c), the degree of all the terms is not equal.
But in the equation y2 dx + (x2 − xy − y2) dy = 0, the degree of all the terms is 2.
Thus, (d) contains a homogeneous differential equation.
Q50.
Answer :
c) 1x
We have,
xdydx-y=2×2⇒dydx-1xy=2xComparing with dydx+Py=Q, we getP=-1x Q=2xNow,I.F.=e-∫1xdy =e-logx=elog1x=1x
Q51.
Answer :
d) 11-y2
We have,1-y2dxdy+yx=aydxdy+y1-y2 x=ay1-y2Comparing with dxdy+Px=Q, we getP=y1-y2 Q=ay1-y2Now, I.F.=e∫y1-y2dy=e-12∫-2y1-y2dy=e-12log1-y2=elog11-y2=11-y2
Q52.
Answer :
(c) y = Cx
We have,y dx-x dyy=0⇒y dx=x dy⇒1ydy=1 xdxIntegrating both sides, we get∫1ydy=∫1 xdx⇒log y=logx+D⇒log y-log x=log C⇒logyx=log C⇒yx=C⇒y=Cx
Q53.
Answer :
c xe∫P1dy=∫Q1e∫P1dydy+C
We have,
dxdy+P1x=Q1
Comparing with the equation dxdy+Px=Q, we get
P = P1
Q = Q1
The general solution of the equation dxdy+Px=Q is given by
xe∫Pdy=∫Qe∫Pdydy+C …(1)
Putting the value of P and Q in (1), we get
xe∫P1dy=∫Q1e∫P1dydy+C
Q54.
Answer :
(c) y ex + x2 = C
We have,
ex dy + (yex + 2x) dx = 0
Dividing both sides by exdx, we getdydx+y+2xex=0⇒dydx+y=-2xexComparing with dydx+Py=Q, we getP=1Q=-2xexNow, I.F.=e∫dx =exSolution is given by, y×I.F.=∫Q×I.F. dx+C ⇒yex=-∫ex×2xexdx+C⇒yex=-2∫x dx+C⇒yex=-x2+C⇒yex+x2=C
Page 22.138 (Revision Exercise)
Q1.
Answer :
i dsdt4+3sd2sdt2=0
The highest order derivative in the given equation is d2sdt2 and its power is 1.
Therefore, the given differential equation is of second order and first degree.
i.e., Order = 2 and degree = 1
(ii) y”‘ + 2y” + y’ = 0
The highest order derivative in the given equation is y”’ and its power is 1.
Therefore, the given differential equation is of third order and first degree.
i.e., Order = 3 and degree = 1
(iii) (y”‘)2 + (y”)3 + (y’)4 + y5 = 0
The highest order derivative in the given equation is y”’ and its power is 2.
Therefore, the given differential equation is of third order and second degree.
i.e., Order = 3 and degree = 2
(iv) y”‘ + 2y” + y’ = 0
The highest order derivative in the given equation is y”’ and its power is 1.
Therefore, the given differential equation is of third order and first degree.
i.e., Order = 3 and degree = 1
(v) y” + (y’)2 + 2y = 0
The highest order derivative in the given equation is y” and its power is 1.
Therefore, the given differential equation is of second order and first degree.
i.e., Order = 2 and degree = 1
(vi) y” + 2y’ + sin y = 0
The highest order derivative in the given equation is y” and its power is 1.
Therefore, the given differential equation is of second order and first degree.
i.e., Order = 2 and degree = 1
(vii) y”‘ + y2 + ey’ = 0
The highest order derivative in the given equation is y”’ and its power is 1.
Therefore, the given differential equation is of third order. This equation cannot be expressed as a polynomial of derivative.
Thus, the degree is not defined.
i.e., Order = 3 and degree is not defined.
Q2.
Answer :
We have,d2ydx2+dydx-6y = 0 …..1Now,y=e-3x⇒dydx=-3e-3x⇒d2ydx2=9e-3x
Putting the values of d2ydx2, dydx and y in 1, we get
LHS=9e-3x-3e-3x-6e-3x = 0 =RHS
Thus, y = e-3x is the solution of the given differential equation.
Q3.
Answer :
(i) We have,
y” − y’ = 0 …..(1)
Now,
y = ex +1
⇒y’=ex⇒y”=ex
Putting the above values in (1), we get
LHS=ex-ex =0=RHS
Thus, y = ex +1 is the solution of the given differential equation.
(ii) We have,
y’ − 2x − 2 = 0 …..(1)
Now,
y = x2 + 2x + C
⇒y’=2x+2
Putting the above value in (1), we get
LHS=2x+2-2x-2=0=RHS
Thus, y = x2 + 2x + C is the solution of the given differential equation.
(iii) We have,
y’ + sin x = 0 …..(1)
Now,
y = cos x + C
⇒y’=-sin x
Putting the above value in (1), we get
LHS=-sin x+sin x=0=RHS
Thus, y = cos x + C is the solution of the given differential equation.
(iv) We have,
y’ = xy1+x2 …..(1)
Now,
y = 1+x2
⇒y’=x1+x2
Putting the above value in (1), we get
LHS=x1+x2=x1+x2×1+x21+x2=xy1+x2=RHS
Thus, y = 1+x2 is the solution of the given differential equation.
(v) We have,
xy’ = y + x x2-y2 …..(1)
Now,
y = x sin x
⇒y’=sin x+x cosx
Putting the above value in (1), we get
LHS=xsin x+xcos x =xsin x+x2cosx=xsin x+xx cos x=xsin x+xx1-sin2 x=xsin x+xx2-x2sin2 x=y+xx2-y2=RHS
Thus, y = x sin x is the solution of the given differential equation.
(v) We have,
xy’ = y + x x2-y2 …..(1)
Now,
y = x sin x
⇒y’=sin x+x cosx
Putting the above value in (1), we get
LHS=xsin x+xcos x =xsin x+x2cosx=xsin x+xx cos x=xsin x+xx1-sin2 x=xsin x+xx2-x2sin2 x=y+xx2-y2=RHS
Thus, y = x sin x is the solution of the given differential equation.
(vi) We have,
x+ydydx=0 …..(1)
Now,
y=a2-x2
⇒y’=-xa2-x2
Putting the above value in (1), we get
LHS=x+y-xa2-x2 =x+a2-x2-xa2-x2=x+a2-x2-xa2-x2=x-x=0=RHS
Thus, y=a2-x2 is the solution of the given differential equation.
Q4.
Answer :
We have,
y = mx (1)
Differentiating both sides, we get
dydx=m⇒dydx=yx From 1⇒xdydx=y⇒xdydx-y=0
Q5.
Answer :
We have,
y = a sin (x + b) …..(2)
Differentiating both sides, we get
dydx=a cosx+b⇒d2ydx2=-a sinx+b ⇒d2ydx2=-a×ya Using 2⇒d2ydx2=-y ⇒d2ydx2+y=0
Q6.
Answer :
The equation of the parabola having vertex at origin and axis along the positive direction of x-axis is given by
y2 =4ax …..(1)
Since there is only one parameter, so we differentiate it only once.
Differentiating with respect to x, we get
2ydydx=4a
Substituting the value of 4a in (1), we get
y2=2ydydx×x⇒y2=2xydydx⇒y2-2xydydx=0
Q7.
Answer :
The equation of the family of circles with radius 3 units, having its centre on y-axis, is given by
x2+y-a2=32 …..1
Here, a is any arbitrary constant.
Since this equation has only one arbitrary constant, we get a first order differential equation.
Differentiating (1) with respect to x, we get
2x+2y-adydx=0⇒x+y-adydx=0⇒x=a-ydydx⇒xdydx=a-y⇒a=y+xdydx
Substituting the value of a in (1), we get
x2+y-y-xdydx2=32⇒x2+x2dydx2=9⇒x2dydx2+x2=9dydx2⇒x2dydx2-9dydx2+x2=0⇒x2-9dydx2+x2=0⇒x2-9y’2+x2=0Hence, x2-9y’2+x2=0 is the required differential equation.
Q8.
Answer :
The equation of the parabola having vertex at origin and axis along the positive direction of y-axis is given by
x2 =4ay …..(1)
Since there is only one parameter, so we differentiate it only once.
Differentiating with respect to x, we get
2x=4ay’⇒4a=2xy’
Substituting the value of 4a in (1), we get
x2=2xy’×y⇒xy’=2y⇒xy’-2y=0
Q9.
Answer :
The equation of the ellipses having foci on y-axis and centre at the origin is given by
x2a2+y2b2=1 …..(1)
Here,
b > a
Since these are two parameters, so we differentiate the equation twice.
Differentiating with respect to x, we get
2xa2+2yb2y’=0⇒xa2+yb2y’=0 …..2⇒1a2+1b2y’2+yb2y”=0 …..3Multiplying throughout by x, we getxa2+xb2y’2+xyb2y”=0 …..4Subtracting 2 from 4, we get1b2xy’2+xyy”-yy’=0 ⇒xy’2+xyy”-yy’=0
Q10.
Answer :
The equation of the family of hyperbolas having centre at the origin and foci on the X-axis is given by
x2a2-y2b2=1 …..1
Here, a and b are parameters.
Since this equation contains two parameters, so we get a second order differential equation.
Differentiating (1) with respect to x, we get
2xa2-2yb2y’=0 …..2
Differentiating (2) with respect to x, we get
2a2-2b2yy”+y’2=0⇒1a2=1b2yy”+y’2⇒b2a2=yy”+y’2 …..(3)
From (2), we get
2xa2=2yb2y’⇒b2a2=yxy’ …..(4)
From (3) and (4), we get
yxy’=yy”+y’2⇒yy’=xyy”+xy’2Hence, xyy”+xy’2-yy’=0 is the required differential equation.
Q11.
Answer :
We have,xy=aex+be-x+x2Differentiating with respect to x on both sides, we get⇒xdydx+y=aex-be-x+2xAgain differentiating with respect to x on both sides, we get⇒xd2ydx2+dydx+dydx=aex+be-x+2⇒xd2ydx2+2dydx=xy-x2+2 ∵ xy=aex+be-x+x2⇒xd2ydx2+2dydx-xy+x2-2=0
Thus, xy = a ex + b e−x + x2 is the solution of the given differential equation.
Q12.
Answer :
We have,
2dydx2+xdydx-y=0 …..1
Now,
y = C x + 2C2
⇒dydx=CPutting dydx=C and y=Cx+2C2 in 1, we get
LHS=2C2+xC-Cx+2C2=2C2+xC-xC-2C2 =0=RHS
Thus, y = C x + 2C2 is the solution of the given differential equation.
Q13.
Answer :
We have,
x-2ydydx+2x+y=0
Now,y2 − x2 − xy = a
⇒2ydydx-2x-y-xdydx=0⇒2y-xdydx-2x-y=0⇒2y-xdydx=2x+y⇒x-2ydydx=-2x+y⇒x-2ydydx+2x+y=0
Thus, y2 − x2 − xy = a is the solution of the given differential equation.
Q14.
Answer :
We have,
cos xdydx+sin xy=1 …..1
Now,
y = A cos x + sin x
⇒dydx=-A sin x+cos xPutting dydx=-A sin x+cos x and y=A cos x+sin x in 1, we get
LHS=cos x-A sin x+cos x+sin x A cos x+sin x=-A sin x cos x+cos2 x+ A cos x sin x+sin2 x=cos2 x+sin2 x=1=RHS
Thus, y = A cos x + sin x is the solution of the given differential equation.
Q15.
We have,
y = ae2x + be-3x + cex …..(1)
Differentiating with respect to x, we get
dydx=2ae2x-3be-3x+cex …..2⇒d2ydx2=4ae2x+9be-3x+cex⇒d3ydx3=8ae2x-27be-3x+cex⇒d3ydx3=72ae2x-3be-3x+cex-6ae2x+be-3x+cex⇒d3ydx3=7dydx-6y Using 1 and 2⇒d3ydx3-7dydx+6y=0
Page 22.139 (Revision Exercise)
Q16.
Answer :
The equation of the family of parabolas axis parallel to y-axis is given by
x-β2=4ay-α …..(1)
Here, α and β are two arbitrary constants.
Differentiating (1) with respect to x, we get
2x-β=4adydx⇒1=2ad2ydx2⇒0=2ad3ydx3⇒d3ydx3=0
Q17.
Answer :
We have,
x2 + y2 + 2ax + 2by + c = 0 …..(i)
Differentiating (i) with respect to x, we get
2x+2yy’+2a+2by’=0Again differentiating with respect to x, we get2+2y’2+2yy”+2by”=01+y’2+yy”+by”=0b=-1+y’2+yy”y”We have,1+y’2+yy”+by”=0Again differentiating with respect to x, we get2y’y”+y’y”+yy”’+by”’=0On substituting the value of b we get,3y’y”+yy”’+-1+y’2+yy”y”y”’=03y’y”2+yy”y”’-y”’-y’2y”’-yy”’y”=03y’y”2=y”’1+y’2
Q18.
Answer :
We have,dydx=sin3 x cos4 x+xx+1⇒dy=sin3 x cos4 x+xx+1dxIntegrating both sides, we get∫dy=∫sin3 x cos4 x+xx+1dx⇒y=∫sin3 x cos4 x dx +∫xx+1dx ⇒y=I1 +I2 …..1 Here, I1=∫sin3 x cos4 x dxI1=∫xx+1dxNow, I1=∫sin3 x cos4 x dx =∫1-cos2 x cos4x sin x dxPutting t=cos x, we getdt=-sin x dx∴ I1=-∫t4 1-t2dt =∫t6-t4dt =t77-t55+C1 =cos7 x7-cos5 x5+C1 I2=∫xx+1dxPutting t2=x+1, we get2t dt=dx∴I2=2∫t2-1t2dt =2∫t4-t2 dt =2t55-2t33+C2 =2x+1525-2x+1323+C2 Putting the value of I1 and I2 in 1, we gety=cos7 x7-cos5 x5+C1+2x+1525-2x+1323+C2 y=cos7 x7-cos5 x5+2x+1525-2x+1323+C ∵ C=C1+C2Hence, y=cos7 x7-cos5 x5+2x+1525-2x+1323+C is the solution of the given differential equation.
Q19.
Answer :
We have,dydx=1×2+4x+5⇒dydx=1×2+4x+4+1⇒dydx=1x+22+12⇒dy=1x+22+12dxIntegrating both sides, we get∫y=∫1x+22+12dx⇒y=tan-1 x+21+C⇒y=tan-1 x+2+C
Q20.
Answer :
We have,dydx=y2+2y+2⇒dydx=y2+2y+1+1⇒dydx=y+12+12⇒1y+12+12dy=dxIntegrating both sides, we get∫1y+12+12dy=∫dx⇒tan-1 y+11+C=x⇒x=tan-1y+1+C
Q21.
Answer :
We have,dydx+4x=ex⇒dydx=ex-4x⇒dy=ex-4xdxIntegrating both sides, we get∫dy=∫ex-4xdx⇒y=ex-2×2+C⇒y+2×2=ex+C
Q22.
Answer :
We have,dydx=x2ex⇒dy=x2ex dxIntegrating both sides, we get∫dy=∫x2IexII dx⇒∫dy=x2∫ex dx-∫ddxx2∫ex dxdx⇒y=x2ex-2∫xex dx⇒y=x2ex-2∫xI exII dx⇒y=x2ex-2x∫ex dx+2∫ddxx∫ex dxdx⇒y=x2ex-2xex+2ex+C⇒y=x2 -2x+2ex+C
Q23.
Answer :
We have, dydx-x sin2 x=1x log x⇒dydx=1x log x+x sin2 x⇒dydx=1x log x+x 21-cos 2x⇒dydx=1x log x+x 2-x 2cos 2xIntegrating both sides, we get∫dy=∫1x log x+x 2-x 2cos 2x dx⇒∫dy=∫1x log xdx+12∫x dx-12∫x cos 2xdx⇒∫dy=∫1x log xdx+12∫x dx-12∫xI×cos 2xII dx ⇒y=log log x+x24-x2∫cos 2xdx+12∫ddxx∫cos 2x dxdx⇒y=log log x+x24-x sin 2×4-cos 2×8+C
Q24.
Answer :
We have,tan2 x+2 tan x+5dydx=21+tan xsec2 x⇒dy=21+tan xsec2 xtan2 x+2 tan x+5 dxIntegrating both sides, we get∫ dy=∫21+tan xsec2 xtan2 x+2 tan x+5 dx …..1Putting tan2 x+2 tan x+5=t∴2 tan x sec2x+ 2sec2x dx=dt⇒21+tan xsec2 x dx=dtTherefore 1 becomes,∫ dy=∫1t dt⇒y=log t+C⇒y=log tan2 x+2 tan x+5+C
Q25.
Answer :
We have,dydx=sin3 x cos2 x+xex⇒dy=sin3 x cos2 x+xexdxIntegrating both sides, we get∫dy=∫sin3 x cos2 x+xexdx⇒y=∫sin3 x cos2 x dx +∫xexdx ⇒y=I1 +I2 …..1 Here, I1=∫sin3x cos2x dxI2=∫xexdxNow, I1=∫sin3 x cos2 x dx=∫1-cos2 x cos2x sin x dxPutting t=cos x, we getdt=-sin x dx∴I1=-∫t2 1-t2dt =∫-t2+t4dt =-t33+t55+C1 =cos5 x5-cos3 x3+C1 I2=∫xexdx =∫xI exII dx =x∫ex dx-∫ddxx∫ex dxdx =xex-ex+C2 =x-1ex+C2Putting the value of I1 and I2 in 1, we gety=cos5 x5-cos3 x3+C1+x-1ex+C2y=cos5 x5-cos3 x3+x-1ex+C, where C=C1+C2
Q26.
Answer :
We have,
tan y dx + tan x dy = 0
⇒tan xdydx=-tan y ⇒cot y dy=-cot x dxIntegrating both sides, we get∫cot y dy=-∫cot x dx⇒log sin y=- log sin x+log C⇒log sin y+ log sin x=log C⇒log sin ysin x=log C⇒sin ysin x=C⇒sin x sin y=C
Q27.
Answer :
We have,
(1 + x) y dx + (1 + y) x dy = 0
dydx=-y1+xx1+y⇒1+yydy=-1+xxdx⇒1y+ydy=-1x+1dxIntegrating both sides, we get∫1y+1dy=-∫1x+1dx⇒∫1ydy+∫dy=-∫1xdx-∫dx⇒logy+y=-logx-x+C⇒logxy+y+x=C⇒x+y+logxy=C
Q28.
Answer :
We have,
x cos2 y dx = y cos2 x dy
⇒y sec2y dy=x sec2x dxIntegrating both sides, we get∫yI sec2yIIdy=∫xI sec2xII dx⇒y∫sec2ydy-∫dydy×∫sec2y dydy=x∫sec2x dx-∫dxdx×∫sec2x dxdx⇒y tan y-∫tan y dy=x tan x-∫tan x dx-C⇒y tan y-log sec y=x tan x-log sec x-C⇒x tan x-y tan y=logsec x-logsec y+C
Q29.
Answer :
We have,
cos y log (sec x + tan x) dx = cos x log (sec y + tan y) dy
⇒logsec y+tan ycos ydy=logsec x+tan xcos xdxIntegrating both sides, we get∫logsec y+tan ycos ydy=∫logsec x+tan xcos xdx …..1Putting logsec y+tan y=t and logsec x+tan x=u⇒sec2y+sec y tan ysec y+tan ydy=dt and sec2x+sec x tan xsec x+tan xdx=du⇒sec y dy=dt and sec x dx=duTherefore, 1 becomes∫t dt=∫u du⇒t22=u22+C⇒logsec y+tan y22=logsec x+tan x22+C⇒logsec y+tan y2=logsec x+tan x2+2C⇒logsec y+tan y2=logsec x+tan x2+k, where k=2C
Q30.
Answer :
We have,
cosec x log ydy+x2y dx=0⇒log yydy=-x2 cosec xdx⇒log yydy=-x2 sin x dxIntegrating both sides, we get∫log yydy=-∫x2 sin x dx …..1Putting log y=t1ydy=dtTherefore, 1 becomes∫t dt=-∫x2 sin x dx⇒t22=-∫x2I sin xII dx⇒12log y2 =-x2 ∫sin x dx-∫ddxx2∫sin x dxdx⇒12log y2 =x2 cos x+2∫x cos x dx⇒12log y2 =x2cos x-2∫xI cos xII dx⇒12log y2 =x2cos x-2x∫cos x dx+2∫ddxx∫cos x dxdx⇒12log y2 =x2cos x-2x sin x-2cos x+C⇒12log y2 =x2-2cos x-2x sin x+C⇒12log y2+2-x2cos x+2x sin x=C
Q31.
Answer :
We have,
1-x2dy+xy dx=xy2dx⇒1-x2dy=xy2dx-xy dx⇒1-x2dy=xy2-ydx⇒1y2-ydy=x1-x2dxIntegrating both sides, we get∫1y2-ydy=∫x1-x2dx⇒∫1y2-y+14-14dy=∫x1-x2dx⇒∫1y-122-122dy=-12∫-2×1-x2dx⇒12×12log y-12-12y-12+12=-12log 1-x2+log C⇒log y-1y=-12log 1-x2+log C⇒2 log y-1y=-log 1-x2+2 log C⇒log y-12y2=-log1-x2+2 log C⇒log y-12y2+log 1-x2=log C2⇒logy-121-x2y2=log C2⇒y-121-x2y2=C2⇒y-121-x2=y2C2
Q32.
Answer :
We have,
dydx=sin x+x cos xy2 log y+1⇒y 2 log y+1dy= sin x+x cos xdxIntegrating both sides, we get2∫yII log y Idy+∫y dy=∫sin x dx+∫x cos x dx⇒2log y∫y dy-2∫ddylog y×∫y dydy+∫y dy=-cos x+x∫ cos x dx-∫dxdx×∫cos x dx⇒y2log y-∫y dy+∫y dy=-cos x+x sin x dx+cos x+C⇒y2log y=x sin x+C
Q33.
Answer :
We have,xe2y-1dy+x2-1eydx=0⇒xe2y-1dy=1-x2eydx⇒e2y-1eydy=1-x2xdx⇒ey-e-ydy=1x-xdxIntegrating both sides, we get∫ey-e-ydy=∫1x-xdx⇒ey+e-y=log x-12×2+C⇒ey+e-y-log x+12×2=C
Q34.
Answer :
We have,
dydx+1=ex+y …..1Let x+y=v⇒1+dydx=dvdx⇒dydx=dvdx-1Then, 1 becomesdvdx-1+1=ev⇒dvdx=ev⇒e-vdv=dxIntegrating both sides, we get∫e-vdv=∫dx⇒-e-v=x+C⇒-1=evx+C⇒-1=x+Cex+y
Q35.
Answer :
We have,dydx=x+y2 …..1Let x+y=v⇒1+dydx=dvdx⇒dydx=dvdx-1Therefore, 1 becomes∴ dvdx-1=v2⇒dvdx=v2+1⇒1v2+1dv=dxIntegrating both sides, we get∫1v2+1dv=∫dx⇒tan-1 v=x+C⇒v=tanx+C⇒x+y=tanx+C
Q36.
Answer :
We have,cos x+ydy=dx⇒dydx=1cosx+y …..1Let x+y=v⇒1+dydx=dvdx⇒dydx=dvdx-1Therefore, 1 becomes∴ dvdx-1=1cos v⇒dvdx=cos v+1cos v⇒cos vcos v+1dv=dxIntegrating both sides, we get∫cos vcos v+1dv=∫dx⇒∫cos v1-cos v1-cos2 vdv=∫dx⇒∫cos v1-cos vsin2 vdv=∫dx⇒∫cot v cosec v-cot2vdv=∫dx⇒∫cot v cosec v-cosec2 v+1dv=∫dx⇒-cosec v+cot v+v=x+C⇒-cosec x+y+cot x+y+x+y=x+C⇒-cosec x+y+cot x+y+y=C⇒cosec x+y-cot x+y=y-C⇒1-cos x+ysin x+y=y-C⇒2 sin2 x+y22 sin x+y2 cos x+y2=y-C⇒sin x+y2cos x+y=y-C⇒tanx+y2=y-C
Q37.
Answer :
We have,dydx+yx=y2x2⇒dydx=yx2-yxPutting y=vx, we getdydx=v+xdvdx∴ v+xdvdx=v2-v⇒xdvdx=v2-2v⇒1v2-2v dv=1xdxIntegrating both sides, we get∫1v2-2v dv=∫1xdx⇒∫1v2-2v+1-1 dv=∫1xdx⇒∫1v-12-12 dv=∫1xdx⇒12log v-1-1v-1+1 =log x+log C⇒log v-2v12 =log Cx⇒log yx-2yx12 =log Cx⇒log y-2xy12 =log Cx⇒y-2xy12=Cx⇒y-2xy =C2x2⇒y-2x=kx2y, where k=C2
Q38.
Answer :
We have,dydx=yx-yxx+yPutting y=vx, we getdydx=v+xdvdx∴ v+xdvdx=vxx-vxxx+vx⇒v+xdvdx=v1-v1+v⇒xdvdx=v1-v1+v-v⇒xdvdx=v-v2-v-v21+v⇒xdvdx=-2v21+v⇒-1+v2v2 dv=1xdxIntegrating both sides, we get-∫1+v2v2 dv=∫1xdx⇒-12∫1v2dv-12∫1vdv=∫1xdx⇒12v-12log v =log x+log C⇒xy-log yx =2log x+2log C⇒xy-log yx =log C2x2⇒xy =log C2x2+log yx⇒xy =log C2xy⇒exy =C2xy⇒exy =k xy, where k=C2
Q39.
Answer :
We have,x+y-1dy=x+ydxdydx=x+yx+y-1Putting x+y=v, we get⇒1+dydx=dvdx⇒dydx=dvdx-1∴ dvdx-1=vv-1⇒dvdx=vv-1+1⇒dvdx=v+v-1v-1⇒dvdx=2v-1v-1⇒v-12v-1 dv=dxIntegrating both sides, we get∫v-12v+1 dv=∫dx⇒12∫2v2v-1dv-∫12v-1dv=∫dx⇒12∫2v-1+12v-1dv-∫12v-1dv=∫dx⇒12∫dv+12∫12v-1dv-∫12v-1dv=∫dx⇒12∫dv-12∫12v-1dv=∫dx⇒12v-14log 2v-1 =x+C⇒12x+y-14log 2x+2y-1 =x+C⇒2x+y-log 2x+2y-1 =4x+4C⇒2x+y-4x-log 2x+2y-1 =4C ⇒2y-x-log 2x+2y-1 =k, where k=4C
Q40.
Answer :
We have,dydx-y cot x=cosec xComparing with dydx+Py=Q, we getP=-cot x Q=cosec xNow,I.F.=e∫-cot x dx=e-log sin x=elog cosec x=cosec xSo, the solution is given byy cosec x=∫cosec x×cosec x dx+C⇒y cosec x=∫cosec2 x dx+C⇒y cosec x=-cot x +C
Q41.
Answer :
We have,dydx-y tan x=-2sin xComparing with dydx+Py=Q, we getP=-tan x Q=-2sin xNow,I.F.=e∫-tan xdx=e-logsec x=elogcos x=cos xSo, the solution is given byy cosx=-∫2sin x cos x dx+C⇒y cosx=-∫sin 2x dx+C∴ y cosx=cos 2×2+C
Q42.
Answer :
We have,dydx-y tan x=ex sec xComparing with dydx+Py=Q, we getP=-tan x Q=ex sec xNow,I.F.=e∫-tan xdx=e-logsec x=elogcos x=cos xSo, the solution is given by y cosx=∫cos x exsec x dx+C⇒y cosx=∫ex dx+C∴ y cosx=ex+C
Q43.
Answer :
We have,dydx-y tan x=exComparing with dydx+Py=Q, we getP=-tan x Q=exNow,I.F.=e∫-tan x dx=e-logsec x=elogcos x=cos xSo, the solution is given byy cos x=∫excos x dx+C⇒y cosx=I+C …..1Where,I=∫exIIcos xI dx …..2⇒ I=cos x∫ex dx-∫ddxcos x∫ex dxdx⇒ I=cos x ex+∫sin x ex dx⇒ I=cos x ex+∫sin xI exII dx⇒ I=cos x ex+sin x∫ex dx-∫ddxsin x∫ex dxdx⇒ I=cos x ex+sin xex -∫cos x ex dx⇒ I=cos x ex+sin xex -I From 2⇒ 2I=cos x ex+sin xex ⇒ I=ex 2cos x+sin x ∴ y cos x=ex 2cos x+sin x +C From 1
Q44.
Answer :
1+y+x2ydx+x+x3dy=0⇒dx+y1+x2dx+x1+x2dy=0⇒dx+1+x2 ydx+xdy=0⇒1+x2 ydx+xdy=-dx⇒ydx+xdy=-11+x2dx⇒ydx+xdy=-dx1+x2
On integrating both side we get,
xy=-∫11+x2dx⇒xy=-tan-1x+c⇒xy+tan-1x=c
Q45.
Answer :
We have,1+x2dy+2y-1dx=0⇒1+x2dy=1-2ydx⇒dy1-2y=11+x2dxIntegrating both sides, we get∫11-2ydy=∫11+x2dx⇒-12log1-2y=tan-1x-log C⇒-log1-2y=2tan-1x-2log C⇒-2tan-1x=-log C+log1-2y⇒-2tan-1x=log 1-2yC⇒e-2tan-1x=1-2yC⇒Ce-2tan-1x=1-2y⇒1-Ce-2tan-1x=2y⇒12-C2e-2tan-1x=y⇒y=12+Ke-2tan-1x, where K=-C2
Q46.
Answer :
We have,y sec2x+y+7tan xdydx=0⇒y sec2x=-y+7tan xdydx⇒-y-7ydy=sec2xtan xdx⇒-1-7ydy=sec2xtan xdxIntegrating both sides, we get∫-1-7ydy=∫sec2xtan xdx⇒-y-7log y=log tan x+log C⇒-y=log tan x+logy7+log C⇒-y=logCy7tan x⇒e-y=Cy7tan x⇒y7tan x=e-yC⇒y7tan x=ke-y, where k=1C
Q47.
Answer :
2ax+x2dydx=a2+2axdydx=a2+2ax2ax+x2=aa+2xx2a+xLet x=2a tan2θ ⇒dx=4a tanθ sec2θ dθdydx=aa+4a tan2θ2a tan2θ 2a 1+tan2θ∫dy=∫a 1+4 tan2θ2 tan2θ 2a sec2θdx∫dy=∫a 1+4 tan2θ2 tan2θ 2a sec2θ4a tanθ sec2θ dθ=∫a 1+4tan2θtanθdθ=a∫1tanθ+4tanθdθy=a∫cotθ+4 tanθ dθy=alog sinθ+4 −log cosθ+cy=alog sinθ−4log cosθ+cAs, x=2a tan2θ⇒tanθ=x2ay=a log sinθcos4θ+c=a log tanθcos3θ+c=a log x2a×x+2a2a3+cy=a logx12(x+2a)324a2+cy+C=a2 log x+3logx+2a where C=c-alog4a2
Q48.
Answer :
Disclaimer: There seems to be error in the given question.
Q49.
Answer :
We have,x2dy+x2-xy+y2dy=0⇒x2dy=xy-x2-y2dy⇒dydx=xy-x2-y2x2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=x2v-x2-x2v2x2⇒v+xdvdx=v-1-v2⇒xdvdx=-1-v2⇒dv1+v2=-1xdxIntegrating both sides, we get∫dv1+v2dv=-∫1xdx⇒tan-1 v=-log x+log C⇒tan-1 yx=logCx⇒etan-1yx=Cx⇒C=x etan-1yx
Q50.
Answer :
We have,y-xdydx=b1+x2dydx⇒y-b=bx2+xdydx⇒1y-bdy=1bx2+xdxIntegrating both sides, we get∫1y-bdy=∫1bx2+xdx⇒∫1y-bdy=1b∫1×2+1bxdx⇒∫1y-bdy=1b∫1×2+1bx+14b2-14b2dx⇒∫1y-bdy=1b∫1x+12b2-12b2dx⇒log y-b=12×12bblog x+12b-12bx+12b+12b+log C⇒log y-b=log bxbx+1+log C⇒y-b=Cbxbx+1⇒Cbx=y-bbx+1⇒x=ky-bbx+1, where k=1bC
Q51.
Answer :
We have,dydx+2y=sin 3x …..1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=2 and Q=sin 3x. ∴ I.F.=e∫P dx =e∫2 dx = e2xMultiplying both sides of 1 by I.F.=e2x, we gete2x dydx+2y=e2xsin 3x ⇒e2xdydx+2e2xy=e2xsin 3xIntegrating both sides with respect to x, we gety e2x=∫e2xsin 3x dx+C⇒y e2x=I+C …..1Where, I=∫e2xsin 3x dx …..2⇒I=e2x∫sin 3x dx-∫de2xdx∫sin 3x dxdx⇒I=-e2xcos 3×3+23∫e2xcos 3x dx⇒I=-e2xcos 3×3+23e2x∫cos 3x dx-∫de2xdx∫cos 3x dxdx⇒I=-e2xcos 3×3+23e2xsin 3×3-23∫e2xsin 3x dx⇒I=-e2xcos 3×3+2 e2xsin 3×9-49∫e2xsin 3x dx⇒I=-e2xcos 3×3+2 e2xsin 3×9-49I Using 2⇒13I9=-e2xcos 3×3+2 e2xsin 3×9⇒I=9132 e2xsin 3×9-e2xcos 3×3⇒I=e2x132 sin 3x-3 cos 3x …..3From 1 and 3, we gety e2x=e2x132 sin 3x-3 cos 3x+C⇒y=31323sin 3x-cos 3x+Ce-2xHence, y=31323sin 3x-cos 3x+Ce-2x is the required solution.
Q52.
Answer :
We have,dydx+y=4x …..1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=1 and Q=4x∴ I.F.=e∫P dx =e∫ dx = exMultiplying both sides of (1) by I.F.=ex, we getex dydx+y=ex4x ⇒exdydx+exy=ex4xIntegrating both sides with respect to x, we gety ex=4∫x ex dx+C⇒y ex=4∫xI exII dx+C⇒y ex=4x∫ex dx-4∫ddxx∫ex dxdx+C⇒y ex=4xex-4∫ex dx+C⇒y ex=4xex-4ex+C⇒y ex=4x-1ex+C⇒y=4x-1+Ce-xHence, y=4x-1+Ce-x is the required solution.
Q53.
Answer :
We have,dydx+5y=cos 4x …..1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=15 and Q=cos 4x.∴ I.F.=e∫P dx =e∫ 5dx = e5xMultiplying both sides of (1) by I.F.=e5x, we gete5x dydx+5y=e5xcos 4x ⇒e5xdydx+5e5xy=e5xcos 4xIntegrating both sides with respect to x, we gety e5x=∫e5xcos 4x dx+C⇒y e5x=I+C …..2Where,I=∫e5xcos 4x dx …..3⇒I=e5x∫cos 4x dx-∫de5xdx∫cos 4x dxdx⇒I=e5xsin 4×4-54∫e5xsin 4x dx⇒I=e5xsin 4×4-54e5x∫sin 4x dx-∫de5xdx∫sin 4x dxdx⇒I=e5xsin 4×4-54-e5xcos 4×4+54∫e5xcos 4x dx⇒I=e5xsin 4×4+5e5xcos 4×16-2516∫e5xcos 4x dx⇒I=e5xsin 4×4+5e5xcos 4×16-2516I From 3⇒4116I=e5xsin 4×4+5e5xcos 4×16⇒4116I=e5x164sin 4x+5cos 4x⇒I=e5x414sin 4x+5cos 4x …..4From 2 and 4 we get⇒y e5x=e5x414sin 4x+5cos 4x+C⇒y=441sin 4x+54cos 4x+Ce-5xHence, y=441sin 4x+54cos 4x+Ce-5x is the required solution.
Q54.
Answer :
We have,xdydx+x cos2yx=ydydx+cos2yx=yx⇒dydx=yx-cos2yxPutting y=vx, we getdydx=v+xdvdx∴ v+xdvdx=v-cos2v⇒xdvdx=-cos2 v⇒sec2 v dv=-1xdxIntegrating both sides, we get∫sec2 v dv=-∫1xdx⇒tan v =-log x+C⇒tan yx =-log x+C
Q55.
Answer :
We have,cos2 xdydx+y=tan x⇒dydx+sec2 xy=tan xsec2 xComparing with dydx+Px=Q, we getP=sec2 x Q=tan xsec2 xNow,I.F.=e∫sec2 x dx=etan xSo, the solution is given byy×etan x=∫tan xsec2 x×etan x dx + C⇒yetan x=I + C …..1Now,I=∫tan xsec2 x×etan x dxPutting t=tan x, we getdt=sec2 x dx∴ I=∫tI×etII dt =t×∫etdt-∫dtdt×∫etdtdt =tet-∫etdt =tet-et⇒I=tan x etan x-etan x=etan xtan x-1Putting the value of I in 1, we getyetan x=etan xtan x-1+ C
Q56.
Answer :
We have,x cos xdydx+y x sin x+cos x=1⇒dydx+tan x+1xy=1x cos xComparing with dydx+Px=Q, we getP=tan x+1x Q=1x cos xNow,I.F.=e∫tan x+1x dx=elog x sec x=x sec xSo, the solution is given byxy sec x=∫sec2 x dx + C⇒xy sec x=tan x+C
Q57.
Answer :
We have,1+y2+x-e-tan-1 ydydx=0⇒dxdy=e-tan-1 y-x1+y2⇒dxdy+x1+y2=e-tan-1 y1+y2
Comparing with dxdy+Px=Q, we getP=11+y2 Q=e-tan-1 y1+y2Now,I.F.=e∫11+y2dy=etan-1 ySo, the solution is given byx×etan-1 y=∫e-tan-1 y1+y2×etan-1 y dy + C⇒x×etan-1 y=∫11+y2 dy + C⇒xetan-1 y=tan-1 y + C
Q58.
Answer :
We have,
y2+x+1ydydx=0
⇒dydx=-y3xy+1⇒dxdy=-xy+1y3⇒dxdy=-xy2-1y3⇒dxdy+xy2=-1y3
Comparing with dxdy+Px=Q, we getP=1y2Q=-1y3Now, I.F.=e∫1y2dy=e-1ySo, the solution is given byx×e-1y=∫-e-1y1y3 dy + C⇒xe-1y=I + C …..1Now,I=∫-e-1y1y3 dyPutting t=1y, we getdt=-1y2dy∴ I=∫tI×e-tII dt =t×∫e-tdt-∫dtdt×∫e-tdtdt =-tet+∫e-tdt =-te-t-e-t∴ I=-1ye-1y-e-1y=-e-1y1+1yPutting the value of I in 1, we getxe-1y=-e-1y1+1y+ C⇒x=-1+1y+ Ce1y
Page 22.140 (Revision Exercise)
Q59.
Answer :
We have,
2cos xdydx+4y sin x=sin 2x
⇒dydx+4ysin x2 cos x=2sin x cos x2 cos x⇒dydx+2y tan x=sin x
Comparing with dydx+Py=Q, we getP=2tan xQ=sin xNow,I.F.=e2∫tan x dx=e2logsec x=sec2 xSo, the solution is given byy×I.F.=∫Q×I.F. dx + C⇒y sec2 x=∫sin x sec2 x dx + C⇒y sec2 x=∫tan x sec x dx + C⇒y sec2 x=sec x + C⇒y=cos x+C cos2x …..1Now, When x=π3, y=0 ∴ 0=cos π3 + C cos2 π3⇒0=12+C14⇒C=-2Putting the value of C in 1, we gety=cos x-2cos2 x
Q60.
Answer :
We have,1+y2dx=tan-1y-xdy⇒dxdy=tan-1 y-x1+y2⇒dxdy+x1+y2=tan-1 y1+y2
Comparing with dxdy+Px=Q, we getP=11+y2 Q=tan-1 y1+y2Now, I.F.=e∫11+y2dy=etan-1 ySo, the solution is given byx×etan-1 y=∫tan-1 y1+y2×etan-1 y dy + C⇒xetan-1 y=I + C …..1Now, I=∫tan-1 y1+y2×etan-1 y dyPutting t=tan-1 y, we getdt=11+y2dy∴ I=∫tI×etII dt =t×∫etdt-∫dtdt×∫etdtdt =tet-∫etdt =tet-et∴ I=tan-1y etan-1 y-etan-1 y =etan-1 ytan-1 y-1Putting the value of I in 1, we getxetan-1 y=etan-1 ytan-1 y-1+ C
Q61.
Answer :
We have,dydx+y tan x=xn cos x
Comparing with dydx+Py=Q, we getP=tan x Q=xn cos xNow, I.F.=e∫tan x dx=elogsec x=sec xSo, the solution is given byy×I.F.=∫Q×I.F. dx + C⇒y sec x=∫xn cos x sec x dx + C⇒y sec x=∫xn dx + C⇒y sec x=xn+1n+1 + C
Q62.
Answer :
We have,dydx=x+12-y⇒2-ydy=x+1dxIntegrating both sides, we get∫2-ydy=∫x+1dx⇒2y-y22=x22+x+C1⇒x22+x+C1-2y+y22=0⇒x2+2x+y2+2C1-4y=0⇒x2+y2+2x-4y+C=0 Where, C=2C1
Q63.
Answer :
We have,dydx=-4xy2⇒1y2dy=-4x dxIntegrating both sides, we get∫1y2dy=-4∫x dx⇒-1y=-2×2+C …..1Now, When x=0, y=1 ∴ -1=0+C⇒C=-1Putting the value of C in 1, we get-1y=-2×2-1⇒1y=2×2+1⇒y=12×2+1
Q64.
Answer :
i) We have,dydx=1-cos x1+cos x⇒dydx=2sin2 x22cos2 x2⇒dydx=tan2 x2⇒dy=tan2 x2dxIntegrating both sides, we get∫dy=∫tan2 x2dx⇒∫dy=∫sec2 x2-1dx⇒y= 2 tan x2-x+C
ii) We have,dydx=4-y2⇒14-y2dy=dxIntegrating both sides, we get∫14-y2dy=∫dx⇒sin-1 y2=x+C⇒y2=sin x+C⇒y=2sin x+C
iii) We have,dydx=1+x21+y2⇒11+y2dy=1+x2dxIntegrating both sides, we get∫11+y2dy=∫1+x2dx⇒tan-1 y=x+x33+C
iv) We have,y log y dx-x dy=0⇒y log y dx=x dy⇒1xdx=1y log ydy⇒1y log ydy=1xdxIntegrating both sides, we get∫1y log ydy=∫1xdx …..1Putting log y=t⇒1ydy=dtTherefore 1 becomes∫1tdt=∫1xdx⇒log t=log x + log C⇒log log y=log x + log C⇒log log y=log Cx⇒log y=Cx⇒y=eCx
v) We have,dydx=sin-1x⇒dy=sin-1xdxIntegrating both sides, we get∫dy=∫sin-1xdx⇒∫dy=∫1II×sin-1xI dx ⇒∫dy=sin-1x∫1 dx-∫ddxsin-1x∫1 dxdx⇒y=x sin-1x-∫x1-x2dxPutting t2=1-x2, we get2t dt=-2x dx⇒-t dt=x dx∴ y=x sin-1x+∫dt⇒y=x sin-1x+t+C⇒y=x sin-1x+1-x2+C
vi) We have,dydx+y=1⇒dydx=1-y⇒11-ydy=dxIntegrating both sides, we get-∫1y-1dy=∫dx⇒∫1y-1dy=-∫dx⇒log y-1=-x+log C⇒log y-1-log C=-x⇒log y-1C=-x⇒y-1C=e-x⇒y=1+Ce-x
Q65.
Answer :
i) We have,xx2-1dydx=1⇒ dydx=1xx2-1⇒dy=1xx2-1dxIntegrating both sides, we get∫dy=∫1xx2-1dx⇒y=∫1xx2-1dx+C⇒y=∫1xx+1x-1dx+C …..1Let 1xx+1x-1=Ax+Bx+1+Cx-1⇒1=Ax+1x-1+Bxx-1+Cxx+1⇒1=Ax2-1+Bx2-x+Cx2+x⇒1=x2A+B+C+x-B+C-AComparing both sides, we get-A=1 …..2-B+C=0 …..3A+B+C=0 …..4Solving 2, 3 and 4, we getA=-1B=12C=12∴1xx+1x-1=-1x+12x+1+12x-1Now, 1 becomesy=∫-1x+12x+1+12x-1dx+C⇒y=-∫1xdx+12∫1x-1dx+12∫1x-1dx⇒y=-log x+12log x-1+12log x+1+C⇒y=12log x-1+12log x+1-log x+CGiven: y2=0∴ 0=12log 2-1+12log 2+1-log 2+C⇒C=log 2-12log 3Substituting the value of C, we gety=12log x-1+12log x+1-log x+log 2-12log 3⇒2y=log x-1+log x+1-2log x+2log 2-log 3⇒2y=log x-1+log x+1-log x2+log 4-log 3⇒2y=logx-1x+1×2-log3-log4⇒y=12logx2-1×2-12log 34
ii) We have,cos dydx=a⇒ dydx=cos-1 a⇒dy=cos-1 a dxIntegrating both sides, we get∫dy=∫cos-1 a dx⇒y=x cos-1 a+CNow, When x=0, y=1 ∴ 1=0+C⇒C=1Putting the value of C in 1, we gety=x cos-1 a+1⇒cosy-1x=a
iii) We have,dydx=y tan x⇒1ydy=tan x dxIntegrating both sides, we get∫1ydy=∫tan x dx⇒log y=log sec x+C ….1Now, When x=0, y=1 ∴ log 1=log 1+C⇒C=0Putting the value of C in 1, we getlog y=log sec x⇒y=sec x
Q66.
Answer :
i) We have,x-ydydx=x+2y⇒dydx=x+2yx-y …..1Clearly this is a homogeneous equation,Putting y=vx⇒dydx=v+xdvdxSubstituting y=vx and dydx=v+xdvdx 1 becomes,v+xdvdx=x+2vxx-vx⇒v+xdvdx=1+2v1-v⇒xdvdx=1+2v1-v-v⇒xdvdx=1+2v-v+v21-v⇒xdvdx=v2+v+11-v⇒1-vv2+v+1dv=1xdx⇒-vv2+v+1+1v2+v+1dv=1xdx⇒-12×2v+1-1v2+v+1+1v2+v+1dv=1xdx⇒-12×2v+1v2+v+1+12×1v2+v+1+1v2+v+1dv=1xdx⇒-12×2v+1v2+v+1+32×1v2+v+1dv=1xdx⇒-12×2v+1v2+v+1+32×1v2+v+14+34dv=1xdx⇒-12×2v+1v2+v+1+32×1v+122+322dv=1xdxIntegrating both sides, we get⇒∫-12×2v+1v2+v+1+32×1v+122+322dv=∫1xdx⇒-12∫2v+1v2+v+1dv+32∫1v+122+322dv=∫1xdx⇒-12log v2+v+1+32×132tan-1v+1232=log x+C⇒-12log yx2+yx+1+32×132tan-1yx+1232=log x+C⇒-12log y2+xy+x2x2+3tan-12y+x3x=log x+C⇒-12log y2+xy+x2+12log x2+3tan-12y+x3x=log x+C⇒-12log y2+xy+x2+log x+3tan-12y+x3x=log x+C⇒-12log y2+xy+x2+3tan-12y+x3x=C⇒log y2+xy+x2-23tan-12y+x3x=-2C⇒log y2+xy+x2=23tan-12y+x3x-2C⇒log y2+xy+x2=23tan-12y+x3x+k Where, k=-2C
ii) We have,x cos yxdydx=y cosyx+x⇒dydx=y cosyx+xx cos yx …..1Clearly this is a homogeneous equation,Putting y=vx⇒dydx=v+xdvdxSubstituting y=vx and dydx=v+xdvdx in 1 we getdydx=y cosyx+xx cos yx⇒v+xdvdx=vx cos v+xx cos v⇒v+xdvdx=v cos v+1cos v⇒xdvdx=v cos v+1cos v-v⇒xdvdx=v cos v+1- v cos vcos v⇒xdvdx=1cos v⇒cos v dv=1xdxIntegrating both sides, we get∫cos v dv=∫1xdx⇒sin v=log x+log C⇒sin yx=log Cx
iii) We have,y dx+x log yxdy-2x dy=0⇒x log yxdy-2x dy=-y dx⇒log yx-2x dy=-y dx⇒dydx=-ylog yx-2x⇒dydx=yx2-log yx …..1Clearly this is a homogenous equation,Putting y=vx⇒dydx=v+xdvdxSubstituting y=vx and dydx=v+xdvdx in 1 we getv+xdvdx=v2-log v⇒xdvdx=v2-log v-v⇒xdvdx=v-2v+v log v2-log v⇒xdvdx=-v+v log v2-log v⇒2-log v-v+v log vdv=1xdx⇒log v-2v log v-vdv=-1xdx⇒log v-1-1v log v-1dv=-1xdx⇒log v-1v log v-1dv-1v log v-1dv=-1xdx⇒1vdv-1v log v-1dv=-1xdxIntegrating both sides we get∫1vdv-∫1v log v-1dv=-∫1xdx⇒log v-I =-log x-log C …..2Where,I=∫1v log v-1dvPuting log v=t1vdv=dt∴I=∫1t-1dt⇒I=log t-1⇒I=log log v-1 …..3From 2 and 3 we getlog v-log log v-1 =-log x-log C⇒log vlog v-1 =-log Cx⇒vlog v-1=1Cx⇒log v-1=vCx⇒log yx-1=Cy
iv) We have,dydx-y=cos xComparing with dydx+Py=Q, we getP=-1 Q=cos xNow, I.F.=e-1∫dx =e-xSolution is given by,y×I.F.=∫cos x×I.F. dx+C⇒ye-x=∫e-x cos x dx+C⇒ye-x=I+C …..1Where,I=∫e-xIIcos x Idx …..2⇒ I=cos x∫e-x dx-∫ddxcos x∫e-x dxdx⇒ I=-cos x e-x-∫sin x e-x dx⇒ I=-cos x e-x-∫sin xI e-xII dx⇒ I=-cos x e-x-sin x∫e-x dx+∫ddxsin x∫e-x dxdx⇒ I=-cos x e-x+sin x e-x -∫cos x e-x dx⇒ I=-cos x e-x+sin xe-x-I Using 2⇒ 2I=-cos x e-x+sin xe-x ⇒ I=12-cos x+sin xe-x …..3From 1 and 3, we get∴ ye-x=sin x-cos xe-x +C⇒y=12sin x-cos x+Cex
v) We have,xdydx+2y=x2⇒dydx+2xy=xComparing with dydx+Py=Q, we getP=2x Q=xNow,I.F.=e2∫1xdx=e2log x=x2So, the solution is given by y×I.F.=∫Q×I.F. dx +C⇒yx2=∫x3 dx+C⇒yx2=x44+C⇒y=x24+Cx-2
vi) We have,dydx+2y=sin xComparing with dydx+Py=Q, we getP=2 Q=sin xNow,I.F.=e2∫dx =e2xSolution is given by,y×I.F.=∫sin x×I.F. dx+C⇒ye2x=I+C …..1Where,I=∫e2xIIsin xIdx …..2⇒ I=sin x∫e2x dx-∫ddxsin x∫e2x dxdx⇒ I=sin x e2x2-12∫cos x e2x dx⇒ I=sin x e2x2-12∫cos xI e2xII dx⇒ I=sin x e2x2-12cos x∫e2x dx+12∫ddxcos x∫e2x dxdx⇒ I=sin x e2x2-14cos x e2x -14∫sin x e2x dx⇒ I=sin x e2x2-14cos x e2x -14I Using 2⇒ I+14I=12sin x e2x-14cos x e2x ⇒ 54I=142sin x e2x-cos xe2x⇒ I=152sin x-cos xe2x …..3Therefore from 1 and 3, we get∴ ye2x=152sin x-cos xe2x +C⇒y=152 sin x-cos x+Ce-2x
vii) We have,dydx+3y=e-2xComparing with dydx+Py=Q, we getP=3Q=e-2xNow,I.F.=e∫P dx=e3∫dx =e3xSo, the solution is given byy×I.F.=∫Q×I.F. dx +C⇒ye3x=∫e3x×e-2xdx+C⇒ye3x=ex+C⇒y=e-2x+Ce-3x
viii) We have,dydx+yx=x2 ⇒dydx+1xy=x2Comparing with dydx+Py=Q, we getP=1x Q=x2Now,I.F.=e∫1xdx =elogx=xSo, the solution is given byy×I.F.=∫Q×I.F. dx +C⇒yx=∫x3+C⇒xy=x44+C
ix) We have,dydx+ sec xy=tan xComparing with dydx+Py=Q, we getP=sec xQ=tan xNow,I.F.=e∫sec x dx=elogsec x+tan x=sec x+tan xSo, the solution is given byy×I.F=∫Q×I.F. dx +C⇒ysec x+tan x=∫sec x+tan xtan x+C⇒ysec x+tan x=∫sec x×tan x dx+∫tan2 x dx+C⇒ysec x+tan x=∫sec x×tan x dx+∫sec2 x-1 dx+C⇒ysec x+tan x=sec x+tan x-x+C
x) We have,xdydx+2y=x2 log xDividing both sides by x, we getdydx+2yx=x log xComparing with dydx+Py=Q, we getP=2x Q=x log xNow,I.F.=e∫Pdx=e∫2xdx=e2logx=x2So, the solution is given byy×I.F.=∫Q×I.F. dx+C⇒x2y=∫x3IIlog xI dx+C⇒x2y=log x∫x3 dx-∫ddxlog x∫x3 dxdx+C⇒x2y=x4log x4-∫x34dx+C⇒x2y=x4log x4-x416+C⇒y=x2log x4-x216+Cx2⇒y=x2164log x-1+Cx-2
xi) We have,x log xdydx+y=2xlog xDividing both sides by x log x, we getdydx+yx log x=2x log xx logx⇒dydx+yx log x=2 x2⇒dydx+1x log xy=2 x2Comparing with dydx+Py=Q, we getP=1x log x Q=2 x2Now, I.F.=e∫Pdx=e∫1x log xdx=eloglog x=log xSo, the solution is given byy×I.F.=∫Q×I.F. dx+C⇒ylog x=2∫1×2×log x dx+C⇒ylog x=I+C …..1Where,I=2∫1x2II log x Idx⇒ I=2log x∫1×2 dx-2∫ddxlog x∫1×2 dxdx⇒ I=-2xlog x+2∫1×2 dx⇒ I=-2xlog x-2x …..2From 1 and 2 we get∴ ylog x=-2xlog x-2x+C⇒ylog x=-2xlog x+1+C
xii) We have,1+x2dy+2xy dx=cot x dx⇒dydx+2×1+x2y=cot x1+x2Comparing with dydx+Py=Q, we getP=2×1+x2Q=cot x1+x2Now,I.F.=e∫2×1+x2dx =elog1+x2=1+x2So, the solution is given byy×I.F.=∫Q×I.F. dx +C⇒y1+x2=∫cot x1+x2×1+x2 dx+C⇒y1+x2=∫cot x dx+C⇒y1+x2=log sin x+C⇒y=1+x2-1log sin x+C1+x2-1
xiii) We have,x+ydydx=1⇒dydx=1x+yLet x+y=v⇒1+dydx=dvdx⇒dydx=dvdx-1∴ dvdx-1=1v⇒dvdx=1v+1⇒dvdx=1+vv⇒v1+vdv=dxIntegrating both sides, we get∫v1+vdv=∫dx⇒∫v+1-11+vdv=∫dx⇒∫dv-∫11+vdv=∫dx⇒v-log v+1=x-log C⇒x+y-log x+y+1=x-log C⇒y-log x+y+1=-log C⇒y=logx+y+1-log C⇒y=logx+y+1C⇒Cey=x+y+1
xiv) We have,y dx+x-y2dy=0⇒y dx=-x-y2dy ⇒dxdy=-1yx-y2 ⇒dxdy+1yx=y …..1Clearly, it is a linear differential equation of the form dxdy+Px=Qwhere P=1y and Q=y∴ I.F.=e∫P dy =e∫1ydy = elog y=yMultiplying both sides of 1 by I.F.=y, we getydxdy+1yx= y×y⇒ydxdy+x=y2Integrating both sides with respect to y, we getxy=∫y2dy+C⇒xy=y33+C⇒x=y23+CyHence, x=y23+Cy is the required solution.
xv) We have,x+3y2dydx=y⇒dxdy=1yx+3y2 ⇒dxdy-1yx=3y …..1Clearly, it is a linear differential equation of the formdxdy+Px=Qwhere P=-1y and Q=3y∴ I.F.=e∫P dy =e-∫1ydy = e-log y=1yMultiplying both sides of (1) by I.F.=1y, we get1ydxdy-1yx= 1y×3y⇒1ydxdy-1yx=3Integrating both sides with respect to y, we getx1y=∫ 3dy+C⇒x1y=3y+C⇒x=3y2+CyHence, x=3y2+Cy is the required solution.
Q67.
Answer :
i) We have,1+x2dydx+2xy=11+x2⇒dydx+2×1+x2y=11+x22Comparing with dydx+Py=Q, we getP=2×1+x2 Q=11+x22Now, I.F.=e∫2×1+x2dx =elog 1+x2=1+x2So, the solution is given byy×I.F.=∫Q×I.F. dx +C⇒y1+x2=∫11+x2 dx+C⇒y1+x2=tan-1 x+C …..1Now, When x=1, y=0 ∴ 01+1=tan-1 1+C⇒C=-1⇒C=-π4Putting the value of C in 1, we gety1+x2=tan-1 x-π4
ii) We have,x+ydy+x-ydx=0dydx=y-xx+yLet y=vxdydx=v+xdvdx∴ v+xdvdx=vx-xx+vx⇒xdvdx=v-11+v-v⇒x dvdx=v-1-v-v21+v⇒xdvdx=-v2+11+v⇒1+vv2+1dv=-1xdx
Integrating both sides, we get∫1+v1+v2dy=-∫1xdx∫112+v2dy+12∫2v1+v2=-∫1xdx⇒tan-1 v+12log1+v2=-log x+C⇒2tan-1 v+log1+v2+2log x =2C⇒2tan-1 v+log1+v2x2=k where, k=2C⇒2tan-1 yx+log1+y2x2x2=k⇒2tan-1 yx+log x2+y2=k …..1Now, When x=1, y=1 ∴ 2tan-1 1+log 2=k⇒k=π2+log 2Putting the value of k in 1, we get2tan-1 yx+log x2+y2=π2+log 2
iii) We have,x2dy+xy+y2dx=0dydx=-xy+y2x2Let y=vx⇒dydx=v+xdvdx∴ v+xdvdx=-vx2+v2x2x2⇒xdvdx=-v+v2-v⇒xdvdx=-2v-v2⇒1v2+2vdv=-1xdx
Integrating both sides, we get∫1v2+2vdy=-∫1xdx∫1v2+2v+1-1dy=-∫1xdx∫1v+12-12dy=-∫1xdx⇒12×1logv+1-1v+1+1=-log x+log C⇒12logvv+2=-log x+log C⇒logvv+2=-2log x+2log C⇒logvv+2+log x2=log C2⇒logvx2v+2=log C2 ⇒vx2v+2=C2 ⇒vx2v+2=k, where k=2C⇒yxx2yx+2=k ⇒x2yy+2x=k …..1Now, When x=1, y=1 ∴ 11+2=k ⇒k=13Putting the value of k in 1, we getx2yy+2x=13⇒3x2y=±y+2xBut y1=1 does not satisfy the equation 3x2y=-y+2x. ∴ 3x2y=y+2x⇒y=2x3x2-1
Q68.
Answer :
We have,x dy=2×2+1dx⇒dy=2×2+1xdx⇒dy=2x+1xdxIntegrating both sides, we get∫dy=∫2x+1xdx⇒y=x2+log x+C …..1Now the given curve passes through 1, 1Therefore, when x=1, y=1∴ 1=1+0+C⇒C=0Putting the value of C in 1, we gety=x2+logx
Page 22.141 (Revision Exercise)
Q69.
Answer :
We have,dydx=2xy2⇒y2dy=2x dxIntegrating both sides, we get∫y2dy=2∫x dx⇒y33=x2+C …..1Now the given curve passes theough -2, 3Therefore, when x=-2, y=3 Substituting x=-2 and y=3 in 1 we get333=-22+C⇒ 9=4+C⇒C=5Putting the value of C in 1, we gety33=x2+5⇒y3=3×2+15⇒y=3×2+1513
Q70.
Answer :
We have,
dydx=ex sin x
⇒dy=exsin x dxIntegrating both sides, we get∫dy=∫exsin x dx⇒y=I+C …..1Where I=∫exIIsin x Idx+C …..2⇒ I=sin x∫ex dx-∫ddxsin x∫ex dxdx⇒ I=sin x ex-∫cos x ex dx⇒ I=sin x ex-∫cos xI exII dx⇒ I=sin x ex-cos x∫ex dx+∫ddxcos x∫ex dxdx⇒ I=sin x ex-cos x ex -∫sin x ex dx⇒ I=sin x ex-cos x ex -I From 2⇒ 2I=sin x ex-cos x ex⇒ I=12exsin x-cos x …..3From 1 and 3 we get∴ y=12exsin x-cos x+C …4Now equation of the curve passes through 0, 0Therefore when x=0; y=0Putting x=0 and y=0 in 4 we get∴ 0=12e0sin 0-cos 0+C⇒C=12Substituting the value of C in 4, we gety=12exsin x-cos x+12⇒2y-1=exsin x-cos x
Q71.
Answer :
The slope of the line having points (x, y) and (−4, −3) is given by y+3x+4.
According to the question,
dydx=2y+3x+4
⇒1y+3dy=2x+4dxIntegrating both sides, we get∫1y+3dy=2∫1x+4dx⇒log y+3=2log x+4+log C⇒log y+3=log Cx+42⇒y+3=Cx+42Since the curve passes through -2, 1, it satisfies the equation of the curve.∴ 1+3=C-2+42⇒C=1Putting the value of C in the equation of the curve, we gety+3=x+42
Q72.
Answer :
We have,
dydx=x2+y22xyLet y=vx⇒dydx=v+xdvdx∴ v+xdvdx=x2+v2x22vx2⇒xdvdx=1+v22v-v⇒xdvdx=1+v2-2v22v⇒xdvdx=1-v22v⇒2v1-v2dv=1xdx
Integrating both sides, we get∫2v1-v2dy=∫1xdx⇒-log 1-v2=log x-log C⇒log 1-v2C=-log x⇒1-v2=Cx⇒1-yx2=Cx⇒x2-y2x2=Cx⇒x2-y2=Cx
Q73.
Answer :
According to the question,
dydx=x+xy⇒dydx=x1+y⇒1y+1dy=x dxIntegrating both sides, we get∫1y+1dy=∫x dx⇒log y+1=x22+log C⇒log y+1C=x22⇒y+1=Cex22Since, the curve passes through 0, 1.It satisfies the equation of the curve.∴1+1=Ce0⇒C=2Puting the value of C in the equation of the curve. We get y+1=2ex22⇒ y=-1+2ex22
Q74.
Answer :
According to the question,
dydx=x+y
⇒dydx-y=x
Comparing with dydx+Py=Q, we getP=-1 Q=xNow, I.F.=e-∫dx =e-xSo, the solution is given byy×I.F.=∫Q×I.F. dx +C⇒ye-x=∫xIe-xIIdx+C⇒ye-x=x∫e-x dx-∫ddxx∫e-x dxdx+C⇒ye-x=-xe-x+∫e-x dx+C⇒ye-x=-xe-x-e-x+CSince the curve passes throught the origin, it satisfies the equation of the curve.⇒0e0=-0e0-e0+CC=1Putting the value of C in the equation of the curve, we getye-x=-xe-x-e-x+1⇒ye-x+xe-x+e-x=1⇒y+x+1e-x=1⇒x+y+1=ex
Q75.
Answer :
According to the question,
dydx+5=x+y
⇒dydx-y=x-5
Comparing with dydx+Py=Q, we getP=-1 Q=x-5Now,I.F.=e-∫dx =e-x So, the solution is given byy×I.F.=∫Q×I.F. dx +C⇒ye-x=∫x-5e-xdx +C⇒ye-x=∫xIe-xIIdx-5∫e-x dx+C⇒ye-x=x∫e-x dx-∫ddxx∫e-x dxdx+5e-x+C⇒ye-x=-xe-x+∫e-x dx+5e-x+C⇒ye-x=-xe-x-e-x+5e-x+C⇒ye-x=-xe-x+4e-x+CSince the curve passes throught the point 0, 2, it satisfies the equation of the curve.2e0=-0e0+4e0+C⇒C=-2Putting the value of C in the equation of the curve, we getye-x=-xe-x+4e-x-2⇒y=-x+4-2ex⇒y=4-x-2ex
Q76.
Answer :
According to the question,
dydx=12y⇒2y dy=dxIntegrating both sides, we get2∫y dy=∫dx⇒y2=x+CSince the curve passes throught the point 4, 3, it satisfies the equation of the curve.9=4+C⇒C=5Putting the value of C in the equation of the curve, we gety2=x+5
Q77.
Answer :
Let N be the initial amount of radium and P be the amount of radium present at any time t.
We have,dPdtαP⇒dPdt=aP, where a<0⇒dPP=adt⇒log P=at+C …..1Now, P=N at t=0Putting P=N and t=0 in 1, we getlog N=C Putting C=logN in 1, we getlog P=at+log N⇒log PN=at …..2According to the question,log2NN=at⇒log2=at⇒t=1alog2, where a is a constant of proportionality
Q78.
Answer :
Let the original amount of radium be N and the amount of radium at any time t be P.
We have,dPdtαP⇒dPdt=-aP⇒dPP=-a dt⇒log P=-at+C …..1Now, P=N at t=0 Putting P=N and t=0 in 1, we getlog N=CPutting C=log N in 1, we getlog P=-at+log N⇒logPN=-at …..2According to the question,P=12N at t=1590logN2N=-1590a⇒a=11590log 2Putting a=11590log2 in 2, we getlogPN=-11590log2tPN=e-log 21590t …..3Putting t=1 in 3 to find the bacteria after 1 year, we getPN=e-log 21590 ⇒PN=0.9996⇒P=0.9996NPercentage of amount disapeared in 1 year=N-PN×100%=N-0.9996NN×100%=0.04%
Q79.
Answer :
Let the original amount of moisture in the porous substance be N and the amount of moisture in the porous substance at any time t be P.
Given: dPdtαP⇒dPdt=-aP⇒dPP=-a dt⇒log P=-at+C …..1Now, P=N at t=0Putting P=N and t=0 in 1, we getlog N=CPutting C=log N in 1, we getlog P=-at+log N⇒log PN=-at …..2According to the question,P=12N at t=1log N2N=-a⇒a=log 2Putting a=log 2 in 2, we getlog PN=-t log2To find the time when it will loss 95% moisture, we haveP=1-95%N=5100N∴ log 5N100N=-t log 2⇒log 20=t log 2⇒t=log 20log 2
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