NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5 - Differential Equations

NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5 - Differential Equations

Edited By Komal Miglani | Updated on May 08, 2025 02:24 PM IST | #CBSE Class 12th

Suppose we have the example of draining out water from a tank, where the drainage rate is directly proportional to the remaining water in the tank. This situation can be represented by a first-order linear differential equation, which can be used to forecast the time taken for the tank to empty, as well as the quantity of water left after a specified time. Prior to this exercise, the NCERT book has already presented solved examples that lead students through the step-by-step solving of such equations—so it is simple to solve the problems of NCERT Exercise 9.5.

The NCERT solutions, designed by experienced teachers, follow the CBSE syllabus for the 2025-26 session and are a precious asset for board exams as well as competitive exams such as the JEE Main. The solutions are presented in a step-by-step and logical format that is easy to comprehend, confidence-building, and enhances analytical ability. Through acquiring these skills, learners are able to achieve proficiency in determining linear differential equations, in solving them using the integrating factor method and comprehending general and specific solutions correctly.

This Story also Contains
  1. Class 12 Maths Chapter 9 Exercise 9.5 Solutions: Download PDF
  2. NCERT Solutions Class 12 Maths Chapter 9: Exercise 9.5
  3. Topics covered in Chapter 9 Differential Equation: Exercise 9.5
  4. NCERT Solutions Subject Wise
  5. Subject Wise NCERT Exemplar Solutions

Class 12 Maths Chapter 9 Exercise 9.5 Solutions: Download PDF

This material provides easy solutions to all the questions of Exercise 9.5 of Differential Equations. The PDF can be downloaded by the students for practice and enhancement of the chapter for board and and competitive exams

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NCERT Solutions Class 12 Maths Chapter 9: Exercise 9.5

Question 1: Find the general solution:

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

Answer:

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:

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

Answer:

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

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

Answer:

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.

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

Answer:

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.

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

Answer:

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.

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

Answer:

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.

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

Answer:

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.

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

Answer:

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.

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

Answer:

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.

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

Answer:

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.

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

Answer:

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.

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

Answer:

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.

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

Answer:

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 $x= \frac{\pi}{3}$
$\sec \frac{\pi}{3} = \sec \frac{\pi}{3}+C\\$

$\\ C = - 2$
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.

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

Answer:

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.

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

Answer:

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 $y= 4\sin^3 x-2\sin^2x\\$

Question 16: Find the equation of a curve passing 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.

Answer:

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 $y+x+1=e^x$

Question 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.

Answer:

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 $y=4-x-2e^x$

Question 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) $x$

Answer:

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)

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

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

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

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

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

Answer:

Given equation is
$(1 - y^2)\frac{dx}{dy} + yx = ay \ \ (-1<y<1)$
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)

Topics covered in Chapter 9 Differential Equation: Exercise 9.5

TopicsDescriptionExample
Identifying Linear Differential EquationEquation in the form: $\frac{d y}{d x}+P y=Q$$\frac{d y}{d x}+3 y=6$
Finding the Integrating Factor (IF)Use: $I F=e^{\int P d x}$, where $P$ is the coefficient of $y$$I F=e^{\int 3 d x}=e^{3 x}$
Multiplying both sides with the IFMultiply full equation by IF to simplify into exact derivative$e^{3 x} \frac{d y}{d x}+3 e^{3 x} y=6 e^{3 x}$
Expressing as a derivativeLeft side becomes the derivative of y×IF$\frac{d}{d x}\left(y e^{3 x}\right)=6 e^{3 x}$
Integrating both sidesSolve the equation by integrating both sides$y e^{3 x}=\int 6 e^{3 x} d x=2 e^{3 x}+C$
Finding the General SolutionSolve for y from the integrated equation$y=2+C e^{-3 x}$
Finding Particular Solution (if values given)Use given values to calculate the constant CIf $y=5$ when $x=0: 5=2+C \Rightarrow C=3$

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Frequently Asked Questions (FAQs)

1. How many questions are explained in the NCERT solutions for Class 12 Maths chapter 9 exercise 9.5?

19 questions are given in Exercise 9.5, Class 12 Maths.

2. Which topic is projected in Class 12 Maths chapter 9 exercise 9.5?

Linear differential equations have been discussed through exercise 9.5.

3. How many multiple-choice questions are given in the Class 12th Maths chapter 9 exercise 9.5?

Two questions of 4 choices are given in exercise 9.5.

4. What is the topic discussed after Class 12th Maths chapter 9 exercise 9.5?

No topics are discussed after exercise 9.5.

5. Is there any exercise after 9.5?

Yes, miscellaneous exercise is given after 9.5.

6. Is it mandatory to solve exercise 9.5 Class 12 Maths for board exam?

The concepts covered in NCERT solutions for Class 12 Maths chapter 9 exercise 9.5 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?

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

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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)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

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)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

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

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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

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