Class 12 Ncert Solutions Maths Differential Equations

                                    Class 12 Ncert Solutions Maths Differential Equations

                                                                       Exercise 9.6


Previous Exercises
Exercise 9.1 &9.2
Exercise 9.4
Exercise 9.5
For each of the differential equations given in Exercises 1 to 12, find the general solution:

1. $\frac{dy}{dx}+ 2y=sinx$

Sol: Comparing the given differential equation with $$\frac{dy}{dx} + Py =Q$$
we get P=2 and Q=sinx
Now, I.F(integrating factor) =$e^{\int P\ dx}=e^{\int 2dx}=e^{2x}$
Hence the solution of the equation is given by
$$\begin{align} y(I.F)&=\int (Q\times I.F)\ dx + C\\ y.e^{2x} &=\int e^{2x}\sin x \ dx +C\tag 1\\ \end{align}$$
Let I = $\int e^{2x}\sin x\ dx$
$\implies$ I=$\sin x\frac{e^{2x}}{2}-\int \cos x\frac{e^{2x}}{2}\ dx$(Integrating by parts)
$\implies$ I=$\frac{1}{2}\sin x e^{2x} - \frac{1}{2}\left[\cos x\frac{e^2x}{2}-\int (-\sin x)\frac{e^2x}{2}\ dx\right]$
$\implies$ I= $\frac{1}{2}\sin x e^{2x} -\frac{1}{4}\cos x e^{2x}-\frac{1}{4}\int \sin x \frac{e^2x}{2}\ dx$
$\implies$ I= $\frac{1}{2}\sin x e^{2x} -\frac{1}{4}\cos x e^{2x}-\frac{I}{4}$
$\implies I+\frac{I}{4} = \frac{1}{4}(2\sin x e^{2x} - \cos x e^{2x})$
$\implies I=\frac{1}{5}(2\sin x-\cos x)e^{2x}$
Substituting the value of I in eq (1) we get
$y.e^{2x}=\frac{1}{5}(2\sin x-\cos x)e^{2x} + C$
$\implies y=\frac{1}{5}(2\sin x -\cos x) + Ce^{-2x}$
This is the required general solution of the given differential equation.

2. $\frac{dy}{dx} +3y=e^{-2x}$

Sol: Comparing the given differential equation with $$\frac{dy}{dx} + Py = Q$$
we get P=3 and Q=e^{-2x}
Thus I.F = $e^{\int P\ dx}=e^{\int 3\ dx}=e^{3x}$
Hence the solution of the equation is
$$\begin{align} y(I.F)&=\int (Q\times I.F)\ dx + C\\ y.e^{3x}&=\int (e^{-2x}\times e^{3x})\ dx + C\\ y.e^{3x}&= \int e^{x}\ dx + C\\ y.e^{3x}&=e^{x} + C\\ y&=\frac{e^x}{e^3x} + \frac{C}{e^{3x}}\\ y&=e^{-2x} + Ce^{-3x}\\ \end{align}$$

3. $\frac{dy}{dx}+\frac{y}{x}=x^2$

Sol: Comparing the given equation with $$\frac{dy}{dx} + Py=Q$$
we get P = $\frac{1}{x}$ and Q=$x^2$
$\therefore$ I.F = $e^{\int \frac{1}{x}\ dx}= e^{\log x}=x$
Hence the solution of the equation is given by
$$\begin{align} y(I.F)&=\int (Q\times I.F)\ dx + C\\ \implies y.x &=\int (x.x^3) + C\\ \implies xy &=\int x^3\ dx + C\\ \implies xy &= \frac{x^4}{4} + C\\ \end{align}$$

4. $\frac{dy}{dx} + (\sec x)y = \tan x$

Sol: Comparing the given equation with $$\frac{dy}{dx} + Py = Q$$
we get P = $\sec x$ and Q = $\tan x$
$\therefore$ I.F. =$e^{\int \sec x}= e^{\log |\sec x + \tan x|}= \sec x + \tan x$
Hence the solution of the equation is given by
$$\begin{align} y(I.F)&=\int (Q\times I.F)\ dx  + C\\ y.(\sec x+\tan x) &=\int (\tan x \times(\sec x + \tan x)) \ dx+ C\\ y.(\sec x+\tan x) &=\int \tan x\sec x\ dx + \int \tan^2 x\ dx + C\\ y.(\sec x+\tan x) &=\sec x + \int (\sec^2 x-1)\ dx + C\\ y.(\sec x+\tan x) &=\sec x + \int (\sec^2 x)\ dx-\int 1\ dx + C\\ y.(\sec x+\tan x) &=\sec x + \tan x - x + C\\ \end{align}$$

5. $\cos^2x\frac{dy}{dx} + y =\tan x$

Sol: The given equation can be written as
$$\frac{dy}{dx} + \sec^2x.y = \tan x\sec^2 x$$
Comparing the equation with $$\frac{dy}{dx} + Py = Q$$
we get P =$\sec^2 x$ and Q =$\tan x\sec^2 x$

$\therefore$ I.F. = $e^{\int \sec^2x\ dx}=e^{\tan x}$
Hence the solution of the differential equation is
$$\begin{align} y(I.F)&=\int (Q\times I.F)\ dx + C\\ y.e^{\tan x}&=\int (\tan x\sec^2 x\times e^{\tan x})\ dx + C\\ y.e^{\tan x} &= \int e^{\tan x}\tan x\ d(\tan x) + C\\ y.e^{\tan x} &= \tan x.e^{\tan x} - \int \frac{d}{d(\tan x)}(\tan x)e^{\tan x}\ d(\tan x) + C\\ y.e^{\tan x} &=\tan x.e^{\tan x} - \int e^{\tan x}\ d(\tan x) + C\\ y.e^{\tan x} &=\tan x.e^{\tan x} - e^{\tan x} + C\\ y.e^{\tan x} &=(\tan x -1)\ e^{\tan x} + C\\ y&= \frac{(\tan x-1)e^{\tan x}}{e^{\tan x}} + \frac{C}{e^{\tan x}}\\ y&=(\tan x-1) + Ce^{-\tan x}\\ \end{align}$$

6. $x\frac{dy}{dx} + 2y=x^2\log x$

Sol: The given equation can be written as
$$\frac{dy}{dx} +\frac{2}{x}y=x\log x$$

Comparing with the equation $$\frac{dy}{dx} +Py = Q$$
we get P =$\frac{2}{x}$ and Q=$x\log x$
$\therefore$ I.F = $e^{\int P\ dx}= e^{\frac{2}{x}\ dx}=e^{2\log x}=e^{\log x^2}= x^2$
Hence the solution of the differential equation is
$$\begin{align} y(I.F)&=\int (Q\times I.F)\ dx + C\\ y.x^2 &=\int x^2(x\ \log x)\ dx  + C\\ y.x^2 &= \int x^3\log x\ dx + C\\ y.x^2 &=\log x. \frac{x^4}{4} - \int \frac{1}{x}.\frac{x^4}{4}\ dx + C(\text{integrating by parts})\\ y.x^2 &= \log x.\frac{x^4}{4} - \frac{1}{4}\int x^3\ dx + C\\ y.x^2 &= \log x.\frac{x^4}{4} - \frac{1}{4}\frac {x^4}{4} + C\\ y.x^2 &= \log x. \frac{x^4}{4} -\frac{x^4}{16} + C\\ y &=\log x. \frac{x^2}{4} - \frac{x^2}{16} + Cx^{-2}\\ y &= \frac{x^2}{16}(4\log |x|-1) + Cx^-2\\ \end{align}$$

7.  $x\ \log x\ \frac{dy}{dx} + y = \frac{2}{x}\ \log x$

Sol: The given differential equation can be written as
$$\begin{align}  \frac{dy}{dx} + \frac{1}{x\log x}y &=\frac{2}{x}\frac{\log x}{x\log x}\\  \frac{dy}{dx} + \frac{1}{x\log x}y &=\frac{2}{x^2}\\ \end{align}$$
Comparing with the equation $$\frac{dy}{dx}+Py=Q$$
we get P=$\frac{1}{x\log x}$ and Q =$\frac{2}{x^2}$
$\therefore$ I.F. = $e^{\int P\ dx}= e^{\int \frac{1}{x\log x}\ dx}= e^{\int \frac{d(\log x)}{\log x}}= e^{\log|\log x|}=\log x$
Hence the solution of the given differential is
$$\begin{align} y(I.F)&=\int (Q\times I.F)\ dx + C\\ y.\log x &=\int \log x\frac{2}{x^2} + C\\ y.\log x &= 2\int x^{-2}\ \log x\ dx + C\\ y.\log x &= 2\log x.\int x^{-2} - \int \left[\frac{d}{dx}(\log x).\int x^{-2}\ dx\right]\ dx + C(\text{integrating by parts})\\ y.\log x &=2\log x\frac{x^{-1}}{-1} - \int \frac{1}{x}.\frac{x^{-1}}{-1}\ dx + C\\ y.\log x &= -2\frac{1}{x}\log x + \int x^{-2}\ dx + C\\ y.\log x &= -\frac{2\log x}{x} + \frac{x^{-1}}{-1} + C\\ y.\log x &=-\frac{2\log x}{x} -\frac{1}{x} + C\\ y.\log x &=-\frac{2}{x}(\log |x| +1) + C\\ \end{align}$$

8. $(1+x^2)\ dy + 2xy\ dx = \cot x\ dx(x\neq0)$

Sol: The given equation can be written as $$\frac{dy}{dx} +\frac{2x}{1+x^2}y=\frac{\cot x}{1+ x^2}$$
Comparing the above equation with $$\frac{dy}{dx} +Py =Q$$
we get
P=$\frac{2x}{1+x^2}$   and Q=$\frac{\cot x}{1+x^2}$
$\therefore$ I.F. = $e^{\int P\ dx}=e^{\int \frac{2x}{1+x^2}\ dx}=e^{\int \frac{d(1+x^2)}{1+x^2}\ dx}= e^{\log|x^2+1|}=x^2+1$
Hence the solution of the differential equation is
$$\begin{align} y(I.F)&=\int (Q\times I.F)\ dx + C\\ y.(x^2 +1) &=\int (x^2+1)\frac{\cot x}{x^2 +1}\ dx + C\\ y.(x^2+1) &= \int \cot x\ dx + C\\ y.(x^2+1) &= \log|\sin x| + C\\ y &=\frac{\log|\sin x|}{x^2 + 1} + \frac{c}{x^2 + 1}\\ y &=(x^2 + 1)^{-1}\ \log|\sin x| + C (x^2 + 1)^{-1}\\ \end{align}$$


9. $x\frac{dy}{dx} + y - x + xy\cot x =0(x\neq0)$

Sol: The given equation can be written as
$$\begin{align} x\frac{dy}{dx} + y(1+x\cot x)  - x &=0\\ \frac{dy}{dx} + (\frac{1}{x} +\cot x)y - 1 &=0\\ \frac{dy}{dx} + (\frac{1}{x} + \cot x)y = 1\\ \end{align}$$
Comparing with the equation $$\frac{dy}{dx} +Py =Q$$ we get
P = $\frac{1}{x} + \cot x$ and Q =1
$\therefore$ I.F. =$e^{\int P dx}= e^{\int (\frac{1}{x} + \cot x)\ dx}=e^{\log |x| + \log|\sin x|}=e^{\log|x\sin x|}= x\sin x$
Hence the solution of the differential equation is
$$\begin{align} y(I.F)&=\int (Q\times I.F)\ dx + C\\ y.(x\sin x) &= \int x\sin x.1\ dx + C\\ y.(x\sin x) &= \int x\sin x\ dx + C\\ y.(x\sin x) &= x(-\cos x) - \int 1.(-cosx)\ dx + C(\text{integration by parts})\\ y.(x\sin x) &=-x\cos x +\int \cos x\ dx + C\\ y.(x\sin x) &=-x\sin x + \sin x + C\\ y &=\frac{-x\cos x}{x\sin x} + \frac{\sin x}{x\sin x} + \frac{C}{x\sin x}\\ y &=\frac{-\cos x}{\sin x} + \frac{1}{x} + \frac{C}{x\sin x}\\ y &=-\cot x + \frac{1}{x} + \frac{C}{x\sin x}\\ \end{align}$$

10. $(x+ y)\frac{dy}{dx}=1$

Sol: The given equation can be written in the form
$$\begin{align} \frac{dy}{dx} &=\frac{1}{x+y}\\ \frac{dx}{dy} &=x+y\\ \frac{dx}{dy} -x &=y\\ \end{align}$$

This is a linear differential equation of the type $$\frac{dx}{dy} + Px = Q$$
Comparing the equation we get P = -1 and Q = y
$\therefore$ I.F.  =$e^{\int (-1)\ dx}=e^{-\int dx}=e^{-x}$
Hence the solution of the equation is
$$\begin{align} x.(I.F) &=\int (Q\times I.F) + C\\ x.e^{-x} &=\int y.e^{-x} + C\\ x.e^{-x} &= y\int e^{-x} - \int \left[\frac{d}{dy}y.\int e^{-x}\ dx\right]\ dx + C(\text{integration by parts})\\ x.e^{-x} &= y\frac{e^{-x}}{-1}- \int 1.\frac{e^{-x}}{-1}\ dx + C\\ x.e^{-x} &= -y.e^{-x} + \int e^{-x} + C\\ x.e^{-x} &= -y.e^{-x} + \frac{e^{-x}}{-1} + C\\ x.e^{-x} &= -y.e^{-x} - e^{-x} + C\\ x &=\frac{-y.e^{-x} - e^{-x} + C}{e^{-x}}\\ x &= -y -1 + Ce^x\\ x+y+1 &= Ce^x\\ \end{align}$$

11. $y\ dx + (x^2 - y^2)\ dy=0$

Sol: The given equation can be written as
$$\begin{align} dx +\frac{x}{y}\ dy - y\ dy &=0\\ \frac{dx}{dy}&= \frac{-x}{y} + y\\ \frac{dx}{dy} + \frac{x}{y} &= y\\ \end{align}$$
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$ where P =$\frac{1}{y}$ and Q = y.
$\therefore$ I.F. =$e^{\int P dy}= e^{\int \frac{1}{y}\ dy}=e^{\log y}= y$
Hence the solution of the equation is
$$\begin{align} x.(I.F) &=\int (Q\times I.F) + C\\ x.y &= \int y.y\ dy + C\\ xy &= \int y^2\ dy + C\\ xy &= \frac{y^3}{3} + C\\ \end{align}$$

12. $(x+3y^2)\frac{dy}{dx} = y$

Sol: The given equation can be written as
$$\begin{align} \frac{dy}{dx} &= \frac{y}{x+ 3y^2}\\ \frac{dx}{dy} &= \frac{x+3y^2}{y}\\ \frac{dx}{dy} &=\frac{x}{y} + 3y\\ \frac{dx}{dy} - \frac{x}{y} &=3y\\ \end{align}$$
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$
where P =$\frac{-1}{y}$ and Q = 3y.
$\therefore$ I.F. =$e^{\int P\ dy}= e^{\int \frac{-1}{y}\ dy}=e^{-\log y}=e^{\log y^{-1}}=y^{-1}=\frac{1}{y}$
Hence the solution of the equation is
$$\begin{align} x.(I.F) &=\int (Q\times I.F) + C\\ x.\frac{1}{y} &=\int (3y.\frac{1}{y})\ dy + C\\ x.\frac{1}{y} &=\int 3\ dy + C\\ x.\frac{1}{y} &=3y + C\\ x &=3y^2 + yC\\ \end{align}$$

For each of the differential equations given in Exercises 13 to 15, find a particular solution satisfying the given condition:

13. $\frac{dy}{dx} +2y\tan x =\sin x; y=0\ when\ x=\frac{\pi}{3}$

Sol: The given equation is a linear differential equation of type $\frac{dy}d{x} + Py = Q$
Here P = 2$\tan x$ and Q =$\sin x$
$\therefore$ I.F. = $e^{\int 2\tan x\ dx}= e^{2\log|\sec x}= e^{\log|\sec x|^2}=(\sec x)^2$
Hence the general solution of the given equation is
$$\begin{align} y(I.F) &= \int (Q\times I.F) + C\\ y.\sec^2 x &= \int (\sec^2 x)\sin x\ dx + C\\ y.\sec^2 x &= \int \frac{\sin x}{\cos x}\sec x\ dx + C\\ y.\sec^2 x &= \int \tan x\sec x\ dx + C\\ y.\sec^2 x &= \sec x+ C\tag 1\\ \end{align}$$
Given that y =0 when x=$\frac{\pi}{3}$. Putting these values in the above equation we get
$$\begin{align} 0 &= \sec \frac{\pi}{3} + C\\ C &= -2\\ \end{align}$$
Substituting the value of C in eq (1) we get
$$\begin{align} y.\sec^2 x &= \sec x - 2\\ y &=\frac{\sec x - 2}{\sec^2 x}\\ y &= \frac{1}{\sec x} - \frac{2}{\sec^2 x}\\ y &= \cos x - 2\cos^2 x\\ \end{align}$$
This is the required particular solution of  the given differential equation.

14. $(1+x^2)\frac{dy}{dx} +2xy =\frac{1}{1+x^2}; y=0\ when\ x=1$

Sol: The given differential equation can be written as
$$\frac{dy}{dx} + \frac{2xy}{1+x^2} = \frac{1}{(1+x^2)^2}$$
Comparing with the linear differential equation of type $\frac{dy}{dx} +Py=Q$ we get
P= $\frac{2x}{1+x^2}$ and Q = $\frac{1}{1+x^2}$
$\therefore$ I.F. =$e^{\int P\ dx}=e^{\frac{2x}{1+x^2}\ dx}=e^{\int \frac{d(1+x^2)}{1+x^2}\ dx}= e^{\log |x^2 +1|}= x^2 + 1$
Hence the general solution of the given differential equation is
$$\begin{align} y(I.F) &= \int (Q\times I.F) + C\\ y.(1+x^2) &= \int \frac{1}{(1+x^2)^2}.(1+x^2)\ dx +C\\ y.(1+x^2) &= \int \frac{dx}{1+x^2} +C\\ y.(1+x^2) &= \tan^{-1}x +C\tag 1\\ \end{align}$$
given that y=0 when x=1. Putting these values in eq(1) we get
$$\begin{align} 0&= tan^{-1} + C\\ C + \frac{\pi}{4} &=0\\ C &= -\frac{\pi}{4}\\ \end{align}$$
Substituting the value of C in eq(1) we get
$$y(1+x^2) = tan^{-1} x - \frac{\pi}{4}$$
This is the required particular solution of the given differential equation.

15. $\frac{dy}{dx} -3y\cot x =\sin 2x; \ y=2 \ when \ x=\frac{\pi}{2}$

Sol: Comparing the given equation with the linear differential equation $\frac{dy}{dx} + Py = Q$
we get P = $-3\cot x$ and Q=$\sin 2x$
$\therefore$ I.F.=$e^{\int P\ dx}=e^{\int -3\cot x\ dx}= e^{-3\log|\sin x|}= e^{\log|\sin x|^{-3}}=(\sin x)^{-3}=\frac{1}{sin^3x}$
Hence the general solution of the equation is given by
$$\begin{align} y(I.F) &= \int (Q\times I.F) + C\\ y.\frac{1}{\sin^3x}&=\int \sin 2x.\frac{1}{\sin^3x}\ dx + C\\ \\ y.\frac{1}{\sin^3x}&=\int \frac{2\sin x\cos x}{\sin^3x}\ dx + C\\ \\ y.\frac{1}{\sin^3x}&=2\int \frac{\cos x}{\sin^2x}\ dx + C\\ \\ y.\frac{1}{\sin^3x}&=2\int \frac{d(\sin x)}{\sin^2x}\ dx + C\\ \\ y.\frac{1}{\sin^3x}&=2\frac{(\sin x)^{-2+1}}{-2+1} + C\\ \\ y.\frac{1}{\sin^3x}&=2\frac{(\sin x)^{-1}}{-1} + C\\ \\ y.\frac{1}{\sin^3x}&=\frac{-2}{\sin x} + C\tag 1\\ \end{align}$$
given that y =2 when x=$\frac{\pi}{2}$. Putting these values in eq (1) we get
$$\begin{align} 2.\frac{1}{(\sin \frac{\pi}{2})^3} &= \frac{-2}{\sin \frac{\pi}{2}} + C\\ 2 &=-2 + C\\ C &=4 \end{align}$$
Substituting the value of C in eq(1) we get
$$\begin{align} y.\frac{1}{\sin^3x}&=\frac{-2}{\sin x} + 4\\ y &= \frac{-2\sin^3x}{\sin x} + 4\sin^3x\\ y &=4\sin^3x - 2sin^2x \end{align}$$
This is the required particular solution of the given differential equation.


16.  Find the equation of a curve through the origin given that the slope of the tangent to the curve at any point (x,y) is equal to the sum of the coordinates of the point.

Sol: We know that slope of any curve at the point (x,y) is given by $\frac{dy}{dx}$
According to the question
$$\begin{align} \frac{dy}{dx} = x+y\\ \frac{dy}{dx} -y=x\tag 1\\ \end{align}$$
Comparing with the linear differential equation of type $\frac{dy}{dx} + Py = Q$ we get
P= -1 and Q =x
$\therefore$ I.F. =$e^{\int(-1dx)}=e^{-x}$
Hence the solution of eq (1) is given by
$$\begin{align} y(I.F) &= \int (Q\times I.F) + C\\ y.e^{-x} &= \int x.e^{-x} + C\\ y.e^{-x} &= x\int e^{-x} - \int \left[\frac{d}{dx}x.\int e^{-x}dx\right]dx+ C(\text{integration by parts})\\ y.e^{-x} &= x\frac{e^{-x}}{-1} - \int 1.\frac{e^{-x}}{-1}dx+ C\\ y.e^{-x} &=- xe^{-x} + \int e^{-x}dx+ C\\ y.e^{-x} &=- xe^{-x} - e^{-x}+ C\\ y.e^{-x} &= -e^{-x}(x + 1)+ C\\ y &= \frac{ -e^{-x}(x + 1)+ C}{e^{-x}}\\ y &= x + 1 + Ce^x\\ x+y+1 &= Ce^x\tag 2\\ \end{align}$$
given that the curve passes through the origin i.e (0,0). So, y =0 when x=0 . Putting these values in eq(2) we get
$$\begin{align} 0+0+1 &=Ce^0\\ C&=1\\ \end{align}$$
Substituting the value of C in eq (2) we get
$$x+y+1=e^x$$
This is the required equation of the curve.

17. Find the equation of a curve passing through the point (0,2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Sol: Let (x,y) be any point on the curve.
According to the question
$$\begin{align} \frac{dy}{dx} + 5= x+y\\ \frac{dy}{dx} -y=x-5\tag 1\\ \end{align}$$
Comparing with the linear differential equation of type $\frac{dy}{dx} + Py = Q$ we get
P= -1 and Q =x-5
$\therefore$ I.F. =$e^{\int(-1dx)}=e^{-x}$
Hence the solution of eq (1) is given by
$$\begin{align} y(I.F) &= \int (Q\times I.F) + C\\ y.e^{-x} &= \int (x-5).e^{-x} + C\\ y.e^{-x} &= \int x.e^{-x}-\int 5e^{-x} + C\\ y.e^{-x} &= x\int e^{-x} - \int \left[\frac{d}{dx}x.\int e^{-x}dx\right]dx- 5\frac{e^{-x}}{-1}+C(\text{integration by parts})\\ y.e^{-x} &= x\frac{e^{-x}}{-1} - \int 1.\frac{e^{-x}}{-1}dx+ 5e^{-x}+ C\\ y.e^{-x} &=- xe^{-x} + \int e^{-x}dx+ 5e^{-x}+C\\ y.e^{-x} &=- xe^{-x} - e^{-x}+5e^{-x}+ C\\ y.e^{-x} &= -e^{-x}(x + 1) +5e^{-x}+C\\ y &= \frac{ -e^{-x}(x + 1)+ C+5e^{-x}}{e^{-x}}\\ y &= -(x + 1) + Ce^x + 5\\ x+y+1 &= 5+ Ce^x\tag 2\\ \end{align}$$
given that the curve passes through the origin i.e (0,2). So, y =2 when x=0 . Putting these values in eq(2) we get
$$\begin{align} 0+2+1 &=5+ Ce^0\\ C&=-2\\ \end{align}$$
Substituting the value of C in eq (2) we get
$$\begin{align} x+y+1=5 +(-2)e^x\\ y=4-x-2e^x\\ \end{align}$$
This is the required equation of the curve.

18. The Integrating factor of the differential equation $x\frac{dy}{dx}-y=2x^2$ is

(A) $e^{-x}$   (B) $e^{-y}$     (C) $\frac{1}{x}$     (D) x

Sol: The given equation can be written as $$\frac{dy}{dx} -\frac{1}{x}y=2x$$
Comparing with the linear differential equation of type $\frac{dy}{dx} + Py =Q$ we get
 P = $\frac{-1}{x}$
$\therefore$ I.F. = $e^{\int P\ dx}=e^{\int\frac{-1}{x}\ dx}=e^{-\log x}=e^{\log x^{-1}}=x^{-1}=\frac{1}{x}$
The correct answer is option (C)

19. The Integrating Factor of the differential equation
$(1-y^2)\frac{dy}{dx} + yx = ay(-1<y<1)$ is

(A) $\frac{1}{y^2-1}$     (B) $\frac{1}{\sqrt{y^2-1}}$    (C) $\frac{1}{1-y^2}$    (D) $\frac{1}{\sqrt{1-y^2}}$

Sol: The given equation can be written as $$\frac{dx}{dy} +\frac{y}{1-y^2}x=\frac{ay}{1-y^2}$$
Comparing with the linear differential equation of type $\frac{dx}{dy} + Px =Q$ we get
 P = $\frac{y}{1-y^2}$
$\therefore$ I.F. = $e^{\int P\ dx}=e^{\int\frac{y}{1-y^2}\ dx}=e^{-\frac{1}{2}\int\frac{2y}{y^2-1}\ dx}=e^{-\frac{1}{2}\log|y^2-1|}=e^{\log|y^2-1|^{-\frac{1}{2}}}=(y^2-1)^{-\frac{1}{2}}=\frac{1}{\sqrt{y^2-1}}$
The correct answer is option (B)


                                       Class 12 Ncert Solutions Maths Differential Equations

Previous Exercises
Exercise 9.1 &9.2

Exercise 9.4
Exercise 9.5

Other Chapters
Chapter 3 Matrices
Chapter 11 Three dimensional geometry

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