NCERT Solutions for Exercise 9.6 Class 12 Maths Chapter 9- Differential Equations

# NCERT Solutions for Exercise 9.6 Class 12 Maths Chapter 9- Differential Equations

Edited By Ramraj Saini | Updated on Dec 04, 2023 08:31 AM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.6

NCERT Solutions for Exercise 9.6 Class 12 Maths Chapter 9 Differential Equations are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for exercise 9.6 Class 12 Maths chapter 9 looks into the questions related to first-order linear differential equations. NCERT solutions for Class 12 Maths chapter 9 exercise 9.6 gives an insight into the concepts of linear differential equations and steps involved in solving linear differential equations. There are a few solved examples before the exercise 9.6 Class 12 Maths. The given sample questions give an idea about the steps involved in solving. Once these example questions are understood, students can solve Class 12 Maths chapter 9 exercise 9.6. The solutions to the Class 12th Maths chapter 6 exercise 9.6 are designed by expert Mathematics faculties and the same can be used to prepare for board exams that follow NCERT and also for competitive exams like JEE Main.

12th class Maths exercise 9.6 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Differential Equations Class 12 Chapter 9 Exercise: 9.6

Question:1 Find the general solution:

$\frac{dy}{dx} + 2y = \sin x$

Given equation is
$\frac{dy}{dx} + 2y = \sin x$
This is $\frac{dy}{dx} + py = Q$ type where p = 2 and Q = sin x
Now,
$I.F. = e^{\int pdx}= e^{\int 2dx}= e^{2x}$
Now, the solution of given differential equation is given by relation
$Y(I.F.) =\int (Q\times I.F.)dx +C$
$Y(e^{\int 2x }) =\int (\sin x\times e^{\int 2x })dx +C$
Let $I =\int (\sin x\times e^{\int 2x })$
$I = \sin x \int e^{2x}dx- \int \left ( \frac{d(\sin x)}{dx}.\int e^{2x}dx \right )dx\\ \\ I = \sin x.\frac{e^{2x}}{2}- \int \left ( \cos x.\frac{e^{2x}}{2} \right )\\ \\ I = \sin x. \frac{e^{2x}}{2}-\frac{1}{2}\left ( \cos x\int e^{2x}dx- \left ( \frac{d(\cos x)}{dx}.\int e^{2x}dx \right ) \right )dx\\ \\ I = \sin x\frac{e^{2x}}{2}-\frac{1}{2}\left ( \cos x.\frac{e^{2x}}{2}+ \int \left ( \sin x.\frac{e^{2x}}{2} \right ) \right )\\ \\ I = \sin x\frac{e^{2x}}{2}-\frac{1}{2}\left ( \cos x.\frac{e^{2x}}{2}+\frac{I}{2} \right ) \ \ \ \ \ \ \ \ \ \ \ (\because I = \int \sin xe^{2x})\\ \\ \frac{5I}{4}= \frac{e^{2x}}{4}\left ( 2\sin x-\cos x \right )\\ \\ I = \frac{e^{2x}}{5}\left ( 2\sin x-\cos x \right )$
Put the value of I in our equation
Now, our equation become
$Y.e^{x^2 }= \frac{e^{2x}}{5}\left (2 \sin x-\cos x \right )+C$
$Y= \frac{1}{5}\left (2 \sin x-\cos x \right )+C.e^{-2x}$
Therefore, the general solution is $Y= \frac{1}{5}\left (2 \sin x-\cos x \right )+C.e^{-2x}$

Question:2 Solve for general solution:

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

Given equation is
$\frac{dy}{dx} + 3y = e^{-2x}$
This is $\frac{dy}{dx} + py = Q$ type where p = 3 and $Q = e^{-2x}$
Now,
$I.F. = e^{\int pdx}= e^{\int 3dx}= e^{3x}$
Now, the solution of given differential equation is given by the relation
$Y(I.F.) =\int (Q\times I.F.)dx +C$
$Y(e^{ 3x }) =\int (e^{-2x}\times e^{ 3x })dx +C$
$Y(e^{ 3x }) =\int (e^{x})dx +C\\ Y(e^{3x})= e^x+C\\ Y = e^{-2x}+Ce^{-3x}$
Therefore, the general solution is $Y = e^{-2x}+Ce^{-3x}$

Question:3 Find the general solution

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

Given equation is
$\frac{dy}{dx} + \frac{y}{x} = x^2$
This is $\frac{dy}{dx} + py = Q$ type where $p = \frac{1}{x}$ and $Q = x^2$
Now,
$I.F. = e^{\int pdx}= e^{\int \frac{1}{x}dx}= e^{\log x}= x$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(x) =\int (x^2\times x)dx +C$
$y(x) =\int (x^3)dx +C\\ y.x= \frac{x^4}{4}+C\\$
Therefore, the general solution is $yx =\frac{x^4}{4}+C$

Question:4 Solve for General Solution.

$\frac{dy}{dx} + (\sec x)y = \tan x \ \left(0\leq x < \frac{\pi}{2} \right )$

Given equation is
$\frac{dy}{dx} + (\sec x)y = \tan x \ \left(0\leq x < \frac{\pi}{2} \right )$
This is $\frac{dy}{dx} + py = Q$ type where $p = \sec x$ and $Q = \tan x$
Now,
$I.F. = e^{\int pdx}= e^{\int \sec xdx}= e^{\log |\sec x+ \tan x|}= \sec x+\tan x$ $(\because 0\leq x\leq \frac{\pi}{2} \sec x > 0,\tan x > 0)$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(\sec x+\tan x) =\int ((\sec x+\tan x)\times \tan x)dx +C$
$y(\sec x+ \tan x) =\int (\sec x\tan x+\tan^2 x)dx +C\\y(\sec x+ \tan x) =\sec x+\int (\sec^2x-1)dx +C\\ y(\sec x+ \tan x) = \sec x +\tan x - x+C$
Therefore, the general solution is $y(\sec x+ \tan x) = \sec x +\tan x - x+C$

Question:5 Find the general solution.

$\cos^2 x\frac{dy}{dx} + y = \tan x\left(0\leq x < \frac{\pi}{2} \right )$

Given equation is
$\cos^2 x\frac{dy}{dx} + y = \tan x\left(0\leq x < \frac{\pi}{2} \right )$
we can rewrite it as
$\frac{dy}{dx}+\sec^2x y= \sec^2x\tan x$
This is $\frac{dy}{dx} + py = Q$ where $p = \sec ^2x$ and $Q =\sec^2x \tan x$
Now,
$I.F. = e^{\int pdx}= e^{\int \sec^2 xdx}= e^{\tan x}$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(e^{\tan x}) =\int ((\sec^2 x\tan x)\times e^{\tan x})dx +C$
$ye^{\tan x} =\int \sec^2 x\tan xe^{\tan x}dx+C\\$
take
$e^{\tan x } = t\\ \Rightarrow \sec^2x.e^{\tan x}dx = dt$
$\int t.\log t dt = \log t.\int tdt-\int \left ( \frac{d(\log t)}{dt}.\int tdt \right )dt \\ \\ \int t.\log t dt = \log t . \frac{t^2}{2}- \int (\frac{1}{t}.\frac{t^2}{2})dt\\ \\ \int t.\log t dt = \log t.\frac{t^2}{2}- \int \frac{t}{2}dt\\ \\ \int t.\log t dt = \log t.\frac{t^2}{2}- \frac{t^2}{4}\\ \\ \int t.\log t dt = \frac{t^2}{4}(2\log t -1)$
Now put again $t = e^{\tan x}$
$\int \sec^2x\tan xe^{\tan x}dx = \frac{e^{2\tan x}}{4}(2\tan x-1)$
Put this value in our equation

$ye^{\tan x} =\frac{e^{2\tan x}}{4}(2\tan x-1)+C\\ \\$
Therefore, the general solution is $y =\frac{e^{\tan x}}{4}(2\tan x-1)+Ce^{-\tan x }\\$

Question:6 Solve for General Solution.

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

Given equation is
$x\frac{dy}{dx} + 2y = x^2\log x$
Wr can rewrite it as
$\frac{dy}{dx} +2.\frac{y}{x}= x\log x$
This is $\frac{dy}{dx} + py = Q$ type where $p = \frac{2}{x}$ and $Q = x\log x$
Now,
$I.F. = e^{\int pdx}= e^{\int \frac{2}{x}dx}= e^{2\log x}=e^{\log x^2} = x^2$ $(\because 0\leq x\leq \frac{\pi}{2} \sec x > 0,\tan x > 0)$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(x^2) =\int (x\log x\times x^2)dx +C$
$x^2y = \int x^3\log x+ C$
Let
$I = \int x^3\log x\\ \\ I = \log x\int x^3dx-\int \left ( \frac{d(\log x)}{dx}.\int x^3dx \right )dx\\ \\ I = \log x.\frac{x^4}{4}- \int \left ( \frac{1}{x}.\frac{x^4}{4} \right )dx\\ \\ I = \log x.\frac{x^4}{4}- \int \left ( \frac{x^3}{4} \right )dx\\ \\ I = \log x.\frac{x^4}{4}-\frac{x^4}{16}$
Put this value in our equation
$x^2y =\log x.\frac{x^4}{4}-\frac{x^4}{16}+ C\\ \\ y = \frac{x^2}{16}(4\log x-1)+C.x^{-2}$
Therefore, the general solution is $y = \frac{x^2}{16}(4\log x-1)+C.x^{-2}$

Question:7 Solve for general solutions.

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

Given equation is
$x\log x\frac{dy}{dx} + y = \frac{2}{x}\log x$
we can rewrite it as
$\frac{dy}{dx}+\frac{y}{x\log x}= \frac{2}{x^2}$
This is $\frac{dy}{dx} + py = Q$ type where $p = \frac{1}{x\log x}$ and $Q =\frac{2}{x^2}$
Now,
$I.F. = e^{\int pdx}= e^{\int \frac{1}{x\log x} dx}= e^{\log(\log x)} = \log x$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(\log x) =\int ((\frac{2}{x^2})\times \log x)dx +C$

take
$I=\int ((\frac{2}{x^2})\times \log x)dx$
$I = \log x.\int \frac{2}{x^2}dx-\int \left ( \frac{d(\log x)}{dt}.\int \frac{x^2}{2}dx \right )dx \\ \\ I= -\log x . \frac{2}{x}+ \int (\frac{1}{x}.\frac{2}{x})dx\\ \\ I = -\log x.\frac{2}{x}+ \int \frac{2}{x^2}dx\\ \\I = -\log x.\frac{2}{x}- \frac{2}{x}\\ \\$
Put this value in our equation

$y\log x =-\frac{2}{x}(\log x+1)+C\\ \\$
Therefore, the general solution is $y\log x =-\frac{2}{x}(\log x+1)+C\\ \\$

Question:8 Find the general solution.

$(1 + x^2)dy + 2xydx = \cot x dx\ (x\neq 0)$

Given equation is
$(1 + x^2)dy + 2xydx = \cot x dx\ (x\neq 0)$
we can rewrite it as
$\frac{dy}{dx}+\frac{2xy}{(1+x^2)}= \frac{\cot x}{1+x^2}$
This is $\frac{dy}{dx} + py = Q$ type where $p = \frac{2x}{1+ x^2}$ and $Q =\frac{\cot x}{1+x^2}$
Now,
$I.F. = e^{\int pdx}= e^{\int \frac{2x}{1+ x^2} dx}= e^{\log(1+ x^2)} = 1+x^2$
Now, the solution of the given differential equation is given by the relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(1+x^2) =\int ((\frac{\cot x}{1+x^2})\times (1+ x^2))dx +C$
$y(1+x^2) =\int \cot x dx+C\\ \\ y(1+x^2)= \log |\sin x|+ C\\ \\ y = (1+x^2)^{-1}\log |\sin x|+C(1+x^2)^{-1}$
Therefore, the general solution is $y = (1+x^2)^{-1}\log |\sin x|+C(1+x^2)^{-1}$

Question:9 Solve for general solution.

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

Given equation is
$x\frac{dy}{dx} + y -x +xy \cot x = 0\ (x \neq 0)$
we can rewrite it as
$\frac{dy}{dx}+y.\left ( \frac{1}{x}+\cot x \right )= 1$
This is $\frac{dy}{dx} + py = Q$ type where $p =\left ( \frac{1}{x}+\cot x \right )$ and $Q =1$
Now,
$I.F. = e^{\int pdx}= e^{\int \left ( \frac{1}{x}+\cot x \right ) dx}= e^{\log x +\log |\sin x|} = x.\sin x$
Now, the solution of the given differential equation is given by the relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(x.\sin x) =\int 1\times x\sin xdx +C$
$y(x.\sin x) =\int x\sin xdx +C$
Lets take
$I=\int x\sin xdx \\ \\ I = x .\int \sin xdx-\int \left ( \frac{d(x)}{dx}.\int \sin xdx \right )dx\\ \\ I =- x.\cos x+ \int (\cos x)dx\\ \\ I = -x\cos x+\sin x$
Put this value in our equation
$y(x.\sin x)= -x\cos x+\sin x + C\\ y = -\cot x+\frac{1}{x}+\frac{C}{x\sin x}$
Therefore, the general solution is $y = -\cot x+\frac{1}{x}+\frac{C}{x\sin x}$

Question:10 Find the general solution.

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

Given equation is
$(x+y)\frac{dy}{dx} = 1$
we can rewrite it as
$\frac{dy}{dx} = \frac{1}{x+y}\\ \\ x+ y =\frac{dx}{dy}\\ \\ \frac{dx}{dy}-x=y$
This is $\frac{dx}{dy} + px = Q$ type where $p =-1$ and $Q =y$
Now,
$I.F. = e^{\int pdy}= e^{\int -1 dy}= e^{-y}$
Now, the solution of given differential equation is given by relation
$x(I.F.) =\int (Q\times I.F.)dy +C$
$x(e^{-y}) =\int y\times e^{-y}dy +C$
$xe^{-y}= \int y.e^{-y}dy + C$
Lets take
$I=\int ye^{-y}dy \\ \\ I = y .\int e^{-y}dy-\int \left ( \frac{d(y)}{dy}.\int e^{-y}dy \right )dy\\ \\ I =- y.e^{-y}+ \int e^{-y}dy\\ \\ I = - ye^{-y}-e^{-y}$
Put this value in our equation
$x.e^{-y} = -e^{-y}(y+1)+C\\ x = -(y+1)+Ce^{y}\\ x+y+1=Ce^y$
Therefore, the general solution is $x+y+1=Ce^y$

Question:11 Solve for general solution.

$y dx + (x - y^2)dy = 0$

Given equation is
$y dx + (x - y^2)dy = 0$
we can rewrite it as
$\frac{dx}{dy}+\frac{x}{y}=y$
This is $\frac{dx}{dy} + px = Q$ type where $p =\frac{1}{y}$ and $Q =y$
Now,
$I.F. = e^{\int pdy}= e^{\int \frac{1}{y} dy}= e^{\log y } = y$
Now, the solution of given differential equation is given by relation
$x(I.F.) =\int (Q\times I.F.)dy +C$
$x(y) =\int y\times ydy +C$
$xy= \int y^2dy + C$
$xy = \frac{y^3}{3}+C$
$x = \frac{y^2}{3}+\frac{C}{y}$
Therefore, the general solution is $x = \frac{y^2}{3}+\frac{C}{y}$

Question:12 Find the general solution.

$(x+3y^2)\frac{dy}{dx} = y\ (y > 0)$

Given equation is
$(x+3y^2)\frac{dy}{dx} = y\ (y > 0)$
we can rewrite it as
$\frac{dx}{dy}-\frac{x}{y}= 3y$
This is $\frac{dx}{dy} + px = Q$ type where $p =\frac{-1}{y}$ and $Q =3y$
Now,
$I.F. = e^{\int pdy}= e^{\int \frac{-1}{y} dy}= e^{-\log y } =y^{-1}= \frac{1}{y}$
Now, the solution of given differential equation is given by relation
$x(I.F.) =\int (Q\times I.F.)dy +C$
$x(\frac{1}{y}) =\int 3y\times \frac{1}{y}dy +C$
$\frac{x}{y}= \int 3dy + C$
$\frac{x}{y}= 3y+ C$
$x = 3y^2+Cy$
Therefore, the general solution is $x = 3y^2+Cy$

Question:13 Solve for particular solution.

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

Given equation is
$\frac{dy}{dx} + 2y \tan x = \sin x; \ y = 0 \ when \ x =\frac{\pi}{3}$
This is $\frac{dy}{dx} + py = Q$ type where $p = 2\tan x$ and $Q = \sin x$
Now,
$I.F. = e^{\int pdx}= e^{\int 2\tan xdx}= e^{2\log |\sec x|}= \sec^2 x$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(\sec^2 x) =\int ((\sin x)\times \sec^2 x)dx +C$
$y(\sec^2 x) =\int (\sin \times \frac{1}{\cos x}\times \sec x)dx +C\\ \\ y(\sec^2 x) = \int \tan x\sec xdx+ C\\ \\ y.\sec^2 x= \sec x+C$
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when $x= \frac{\pi}{3}$
at $x= \frac{\pi}{3}$
$0.\sec \frac{\pi}{3} = \sec \frac{\pi}{3}+C\\ \\ C = - 2$
Now,

$y.\sec^2 x= \sec x - 2\\ \frac{y}{\cos ^2x}= \frac{1}{\cos x}- 2\\ y = \cos x- 2\cos ^2 x$
Therefore, the particular solution is $y = \cos x- 2\cos ^2 x$

Question:14 Solve for particular solution.

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

Given equation is
$(1 + x^2)\frac{dy}{dx} + 2xy =\frac{1}{1 + x^2}; \ y = 0 \ when \ x = 1$
we can rewrite it as
$\frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{1}{(1+x^2)^2}$
This is $\frac{dy}{dx} + py = Q$ type where $p =\frac{2x}{1+x^2}$ and $Q = \frac{1}{(1+x^2)^2}$
Now,
$I.F. = e^{\int pdx}= e^{\int \frac{2x}{1+x^2}dx}= e^{\log |1+x^2|}= 1+x^2$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(1+ x^2) =\int (\frac{1}{(1+x^2)^2}\times (1+x^2))dx +C$
$y(1+x^2) =\int \frac{1}{(1+x^2)}dx +C\\ \\ y(1+x^2) = \tan^{-1}x+ C\\ \\$
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x = 1
at x = 1
$0.(1+1^2) = \tan^{-1}1+ C\\ \\ C =- \frac{\pi}{4}$
Now,
$y(1+x^2)= \tan^{-1}x- \frac{\pi}{4}$
Therefore, the particular solution is $y(1+x^2)= \tan^{-1}x- \frac{\pi}{4}$

Question:15 Find the particular solution.

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

Given equation is
$\frac{dy}{dx} - 3y \cot x = \sin 2x;\ y = 2\ when \ x = \frac{\pi}{2}$
This is $\frac{dy}{dx} + py = Q$ type where $p =-3\cot x$ and $Q =\sin 2x$
Now,
$I.F. = e^{\int pdx}= e^{-3\cot xdx}= e^{-3\log|\sin x|}= \sin ^{-3}x= \frac{1}{\sin^3x}$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(\frac{1}{\sin^3 x}) =\int (\sin 2x\times\frac{1}{\sin^3 x})dx +C$
$\frac{y}{\sin^3 x} =\int (2\sin x\cos x\times\frac{1}{\sin^3 x})dx +C$
$\frac{y}{\sin^3 x} =\int (2\times \frac{\cos x}{\sin x}\times\frac{1}{\sin x})dx +C$
$\frac{y}{\sin^3 x} =\int (2\times\cot x\times cosec x)dx +C$
$\frac{y}{\sin^3 x} =-2cosec x +C$
Now, by using boundary conditions we will find the value of C
It is given that y = 2 when $x= \frac{\pi}{2}$
at $x= \frac{\pi}{2}$
$\frac{2}{\sin^3\frac{\pi}{2}} = -2cosec \frac{\pi}{2}+C\\ \\ 2 = -2 +C\\ C = 4$
Now,
$y= 4\sin^3 x-2\sin^2x\\$
Therefore, the particular solution is $y= 4\sin^3 x-2\sin^2x\\$

Let f(x , y) is the curve passing through origin
Then, the slope of tangent to the curve at point (x , y) is given by $\frac{dy}{dx}$
Now, it is given that
$\frac{dy}{dx} = y + x\\ \\ \frac{dy}{dx}-y=x$
It is $\frac{dy}{dx}+py=Q$ type of equation where $p = -1 \ and \ Q = x$
Now,
$I.F. = e^{\int pdx}= e^{\int -1dx }= e^{-x}$
Now,
$y(I.F.)= \int (Q \times I.F. )dx+ C$
$y(e^{-x})= \int (x \times e^{-x} )dx+ C$
Now, Let
$I= \int (x \times e^{-x} )dx \\ \\ I = x.\int e^{-x}dx-\int \left ( \frac{d(x)}{dx}.\int e^{-x}dx \right )dx\\ \\ I = -xe^{-x}+\int e^{-x}dx\\ \\ I = -xe^{-x}-e^{-x}\\ \\ I = -e^{-x}(x+1)$
Put this value in our equation
$ye^{-x}= -e^{-x}(x+1)+C$
Now, by using boundary conditions we will find the value of C
It is given that curve passing through origin i.e. (x , y) = (0 , 0)
$0.e^{-0}= -e^{-0}(0+1)+C\\ \\ C = 1$
Our final equation becomes
$ye^{-x}= -e^{-x}(x+1)+1\\ y+x+1=e^x$
Therefore, the required equation of the curve is $y+x+1=e^x$

Let f(x , y) is the curve passing through point (0 , 2)
Then, the slope of tangent to the curve at point (x , y) is given by $\frac{dy}{dx}$
Now, it is given that
$\frac{dy}{dx} +5= y + x \\ \\ \frac{dy}{dx}-y=x-5$
It is $\frac{dy}{dx}+py=Q$ type of equation where $p = -1 \ and \ Q = x- 5$
Now,
$I.F. = e^{\int pdx}= e^{\int -1dx }= e^{-x}$
Now,
$y(I.F.)= \int (Q \times I.F. )dx+ C$
$y(e^{-x})= \int ((x-5) \times e^{-x} )dx+ C$
Now, Let
$I= \int ((x-5) \times e^{-x} )dx \\ \\ I = (x-5).\int e^{-x}dx-\int \left ( \frac{d(x-5)}{dx}.\int e^{-x}dx \right )dx\\ \\ I = -(x-5)e^{-x}+\int e^{-x}dx\\ \\ I = -xe^{-x}-e^{-x}+5e^{-x}\\ \\ I = -e^{-x}(x-4)$
Put this value in our equation
$ye^{-x}= -e^{-x}(x-4)+C$
Now, by using boundary conditions we will find the value of C
It is given that curve passing through point (0 , 2)
$2.e^{-0}= -e^{-0}(0-4)+C\\ \\ C = -2$
Our final equation becomes
$ye^{-x}= -e^{-x}(x-4)-2\\ y=4-x-2e^x$
Therefore, the required equation of curve is $y=4-x-2e^x$

(A) $e^{-x}$

(B) $e^{-y}$

(C) $\frac{1}{x}$

(D) $x$

Given equation is
$x\frac{dy}{dx} - y = 2x^2$
we can rewrite it as
$\frac{dy}{dx}-\frac{y}{x}= 2x$
Now,
It is $\frac{dy}{dx}+py=Q$ type of equation where $p = \frac{-1}{x} \ and \ Q = 2x$
Now,
$I.F. = e^{\int pdx} = e^{\int \frac{-1}{x}dx}= e^{\int -\log x }= x^{-1}= \frac{1}{x}$
Therefore, the correct answer is (C)

(A) $\frac{1}{{y^2 -1}}$

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

(C) $\frac{1}{{1 - y^2 }}$

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

Given equation is
$(1 - y^2)\frac{dx}{dy} + yx = ay \ \ (-1
we can rewrite it as
$\frac{dx}{dy}+\frac{yx}{1-y^2}= \frac{ay}{1-y^2}$
It is $\frac{dx}{dy}+px= Q$ type of equation where $p = \frac{y}{1-y^2}\ and \ Q = \frac{ay}{1-y^2}$
Now,
$I.F. = e^{\int pdy}= e^{\int \frac{y}{1-y^2}dy}= e^{\frac{\log |1 - y^2|}{-2}}= (1-y^2)^{\frac{-1}{2}}= \frac{1}{\sqrt{1-y^2}}$
Therefore, the correct answer is (D)

## More About NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.6

A few examples are solved in the book just before the exercise 9.6 Class 12 Maths to get an idea of how these types of differential equations are solved. After the examples Class, 12 Maths chapter 9 exercise 9.6 are given and the NCERT solutions for Class 12 Maths chapter 9 exercise 9.6 are given here. There are 19 questions in the NCERT syllabus Class 12th Maths chapter 6 exercise 9.6. A few of them is to find the particular solutions of linear differential equations and the rest to find the general solutions. Two of the given questions are of multiple-choice type.

## Benefits of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.6

• These solutions of the exercise 9.6 Class 12 Maths are useful to get an idea of topic 9.5.3
• Students can expect one question of the types explained in the NCERT solutions for Class 12 Maths chapter 9 exercise 9.6 for the CBSE board exam.
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## Key Features Of NCERT Solutions for Exercise 9.6 Class 12 Maths Chapter 9

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 9.6 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 9.6, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 9.6 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 9.6 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for ex 9.6 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for class 12 maths ex 9.6 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
##### JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

## Subject Wise NCERT Exemplar Solutions

1. What number of questions are explained in the NCERT solutions for Class 12 Maths chapter 9 exercise 9.6?

19 questions are given in the exercise 9.6 Class 12 Maths.

2. Which topic is projected in Class 12 Maths chapter 9 exercise 9.6?

Linear differential equations have been discussed through exercise 9.6.

3. What number of multiple-choice questions are given in the Class 12th Maths chapter 6 exercise 9.6?

Two questions of 4 choices are given in exercise 9.6.

4. What is the topic discussed after Class 12th Maths chapter 6 exercise 9.6?

No topics are discussed after exercise 9.6.

5. Is there any exercise after 9.6?

Yes, miscellaneous exercise is given after 9.6.

6. Is it mandatory to solve exercise 9.6 Class 12 Maths for board exam?

The concepts covered in NCERT solutions for Class 12 Maths chapter 9 exercise 9.6 are important from the chapter differential equations.

7. Can I skip the chapter differential equations for board exams?

No, it is important to go through the chapter as it holds a good weightage for board exams

8. What is the number of practice exercises given in the chapter differential equations?

7 exercises are present in the Class 12 NCERT Maths chapter Differential Equations.

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Yuvan 01 September,2024

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9