Anand Classes brings you a comprehensive free-downloadable PDF of NCERT Solutions for Class 12 Maths Chapter 9 – Differential Equations, Exercise 9.3, focusing on methods for forming and solving first-order, first-degree differential equations, including representing families of curves and eliminating arbitrary constants. These solutions are prepared in accordance with the latest CBSE syllabus for clear understanding and effective revision. Click the print button to download study material and notes.
NCERT Question.1 : Find the general solution of differential equation :
$$\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$$
Solution:
Using Equation : $cos x = 1 -2 sin^2(x/2) = 2 cos^2(x/2) – 1$
$$\frac{dy}{dx}=\frac{2sin^2(x/2)}{2cos^2(x/2)}$$
The given equation can be rewritten as:
$$\frac{dy}{dx}=\tan^2\frac{x}{2}$$
Using the identity $\tan^2 u=\sec^2 u-1$, we have
$$\frac{dy}{dx}=\sec^2\frac{x}{2}-1$$
So
$$dy=\bigg(\sec^2\frac{x}{2}-1\bigg)dx$$
On integrating both sides, we get
$$\int dy=\int\bigg(\sec^2\frac{x}{2}-1\bigg)dx$$
$$y=\int\sec^2\dfrac{x}{2}dx-\int 1dx$$
Since $\int\sec^2\dfrac{x}{2}dx=2\tan\dfrac{x}{2}$, it follows that
$$y=2\tan\frac{x}{2}-x+C$$
Final Answer:
$$\boxed{y=2\tan\frac{x}{2}-x+C}$$
Download notes by Anand Classes — perfect for JEE and CBSE, top-quality study material to help you revise trigonometric identities and differential equations.
NCERT Question.2 : Find the general solution of differential equation :
$$\frac{dy}{dx}=\sqrt{4-y^2}$$
Solution :
Separate variables:
$$\frac{dy}{\sqrt{4-y^2}}=dx.$$
Integrate both sides:
$$\int\frac{dy}{\sqrt{4-y^2}}=\int dx.$$
Using $\displaystyle\int\frac{dy}{\sqrt{a^2-y^2}}=\sin^{-1}\frac{y}{a}+C$ with $a=2$, we get
$$\sin^{-1}\frac{y}{2}=x+C.$$
Solve for $y$:
$$y=2\sin(x+C).$$
Final Answer
$$\boxed{y=2\sin(x+C)}$$
Download notes by Anand Classes — perfect for JEE and CBSE practice, a concise resource for mastering differential equations.
NCERT Question.3 : Find the general solution of differential equation :
$$\frac{dy}{dx}+y=1$$
Solution:
Rewrite the equation as
$$\frac{dy}{dx}=1-y.$$
Separate variables:
$$\frac{dy}{1-y}=dx.$$
Integrate both sides:
$$\int\frac{dy}{1-y}=\int dx.$$
Using the substitution $u=1-y$ (so $du=-dy$), the left integral gives $-\ln|1-y|$. Thus
$$-\ln|1-y|=x+C.$$
Rearrange (replace $-C$ by a new constant $C$):
$$\ln|1-y|=-x+C\quad\Longrightarrow\quad |1-y|=Ce^{-x}.$$
Absorbing the sign into the constant $C$ (allowing $C$ to be any nonzero real constant) yields
$$1-y=Ce^{-x}.$$
Therefore the general solution is
$$\boxed{y=1-Ce^{-x}}$$
where $C$ is an arbitrary constant.
Get complete revision notes from Anand Classes — ideal for JEE & CBSE prep and concise practice material for first-order differential equations.
NCERT Question.4 : Find the general solution of differential equation :
$$\sec^2 x\tan ydx+\sec^2 y\tan xdy=0$$
Solution:
Divide both sides by $\tan x\tan y$ (assuming $\tan x\neq0,\ \tan y\neq0$):
$$\frac{\sec^2 x\tan y}{\tan x\tan y}dx+\frac{\sec^2 y\tan x}{\tan x\tan y}dy=0$$
which simplifies to
$$\frac{\sec^2 x}{\tan x}dx+\frac{\sec^2 y}{\tan y}dy=0.$$
Set $u=\tan x$ so $du=\sec^2 xdx$, and $v=\tan y$ so $dv=\sec^2 ydy$. The equation becomes
$$\frac{du}{u}+\frac{dv}{v}=0.$$
Integrate both sides:
$$\int\frac{du}{u}+\int\frac{dv}{v}=0\quad\Longrightarrow\quad \ln|u|+\ln|v|=C.$$
Substitute back $u=\tan x$, $v=\tan y$:
$$\ln\big|\tan x\big|+\ln\big|\tan y\big|=C
\quad\Longrightarrow\quad \ln\big|\tan x\tan y\big|=C.$$
Exponentiating and absorbing the sign into the constant gives the general implicit solution
Final Answer:
$$\boxed{\tan x\tan y=C}$$
where $C$ is an arbitrary constant.
Download notes by Anand Classes — perfect for JEE and CBSE practice, a concise resource for mastering trigonometric differential equations.
NCERT Question.5 : Find the general solution of differential equation :
$$(e^{x}+e^{-x})dy-(e^{x}-e^{-x})dx=0$$
Solution:
Rewrite the equation as
$$ (e^{x}+e^{-x})dy=(e^{x}-e^{-x})dx $$
Rewrite as
$$\frac{dy}{dx}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}.$$
Notice that the denominator $e^{x}+e^{-x}$ has derivative
$$\frac{d}{dx}\big(e^{x}+e^{-x}\big)=e^{x}-e^{-x},$$
which is exactly the numerator. Hence the integrand is the derivative of a logarithm:
$$\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}=\frac{d}{dx}\ln\big|e^{x}+e^{-x}\big|.$$
Integrate both sides:
$$y=\int\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}dx=\ln\big|e^{x}+e^{-x}\big|+C.$$
Final Answer
$$\boxed{y=\ln\big(e^{x}+e^{-x}\big)+C}$$
(Equivalently $y=\ln\big(2\cosh x\big)+C$, but the form above avoids naming hyperbolic functions.)
Get concise revision notes from Anand Classes — perfect for JEE and CBSE practice and reliable review of first-order differential equations.
NCERT Question.6 : Find the general solution of differential equation :
$$\frac{dy}{dx}=(1+x^2)(1+y^2)$$
Solution :
Rewriting the given equation:
$$\frac{dy}{1+y^2}=(1+x^2),dx$$
Integrate both sides:
$$\int\frac{dy}{1+y^2}=\int(1+x^2)dx$$
Now,
$$\int\frac{dy}{1+y^2}=\tan^{-1}y$$
and
$$\int(1+x^2)dx=x+\frac{x^3}{3}+C$$
Therefore,
$$\tan^{-1}y=x+\frac{x^3}{3}+C.$$
Final Answer
$$\boxed{\tan^{-1}y=x+\frac{x^3}{3}+C}$$
Download detailed step-by-step notes from Anand Classes — ideal for JEE and CBSE students to master integration and differential equations.
NCERT Question.7 : Find the general solution of differential equation :
$$y\ln ydx – xdy = 0$$
Solution :
Rewrite the equation as
$$y\ln ydx = xdy$$
and separate variables:
$$\frac{dy}{y\ln y}=\frac{dx}{x}.$$
Set $u=\ln y$, so $du=\dfrac{dy}{y}$. The left integral becomes
$$\int\frac{dy}{y\ln y}=\int\frac{du}{u}=\ln\big|u\big|=\ln\big|\ln y\big|.$$
Integrating both sides gives
$$\ln\big|\ln y\big|=\ln|x|+C.$$
Exponentiate (set $K=e^{C}>0$) to get
$$\big|\ln y\big|=K|x|.$$
Absorbing sign and absolute-values into an arbitrary constant $C$ (which may be any real number) yields the general relation
$$\ln y = Cx.$$
Solving for $y$,
$$\boxed{y=e^{Cx}}$$
where $C$ is an arbitrary constant.
(Notice the constant solution $y=1$ corresponds to $C=0$ and is included.)
Download concise notes by Anand Classes — perfect for JEE and CBSE practice and a reliable quick-review for first-order differential equations.
NCERT Question.8 : Find the general solution of differential equation :
$$x^{5}\frac{dy}{dx}=-y^{5}$$
Solution:
Rewrite as a separable equation:
$$\frac{dy}{dx}=-\frac{y^{5}}{x^{5}} \quad\Longrightarrow\quad \frac{dy}{y^{5}}=-\frac{dx}{x^{5}}.$$
Integrate both sides:
$$\int y^{-5}dy=\int -x^{-5}dx$$
Compute the integrals:
$$\frac{y^{-4}}{-4}=\frac{x^{-4}}{4}+C$$
Multiply by $4$ and rearrange (absorb constants) to obtain the implicit general solution:
$$x^{-4}+y^{-4}=C$$
Final Answer:
$$\boxed{x^{-4}+y^{-4}=C}$$
Download practice sheets by Anand Classes — perfect for JEE and CBSE revision, concise and reliable differential-equation practice material.
NCERT Question.9 : Find the general solution of differential equation :
$$\frac{dy}{dx}=\sin^{-1}x$$
Solution :
Integrate both sides with respect to $x$:
$$y=\int \sin^{-1}xdx$$
Use integration by parts with $u=\sin^{-1}x, dv=dx$. Then $du=\dfrac{dx}{\sqrt{1-x^{2}}}$ and $v=x$. Thus
$$
y = x\sin^{-1}x-\int x\cdot\frac{dx}{\sqrt{1-x^{2}}}
= x\sin^{-1}x-\int\frac{xdx}{\sqrt{1-x^{2}}}.
$$
Substitute $t=1-x^{2}$ so $dt=-2xdx$, hence $xdx=-\tfrac{1}{2}dt$. Therefore
$$
\int\frac{xdx}{\sqrt{1-x^{2}}}=\int\frac{-\tfrac{1}{2}dt}{\sqrt{t}}=-\sqrt{t}=-\sqrt{1-x^{2}}.
$$
Putting this back gives
$$
y=x\sin^{-1}x-(-\sqrt{1-x^{2}})=x\sin^{-1}x+\sqrt{1-x^{2}}+C.
$$
Final Answer
$$\boxed{y=x\sin^{-1}x+\sqrt{1-x^{2}}+C}$$
Download concise notes by Anand Classes — perfect for JEE and CBSE practice, a clear revision resource for integration and differential equations.
NCERT Question.10 : Find the general solution of differential equation :
$$e^{x}\tan ydx+(1-e^{x})\sec^{2}ydy=0$$
Solution :
Rewrite and group terms with $dy$ on one side:
$$(1-e^{x})\sec^{2}ydy=-e^{x}\tan ydx.$$
Use the substitution $u=\tan y$, so $du=\sec^{2}ydy$. The equation becomes
$$(1-e^{x})du=-e^{x}udx.$$
Separate variables:
$$\frac{du}{u}=-\frac{e^{x}}{1-e^{x}}dx.$$
Integrate both sides. For the right side use $t=1-e^{x}$ so $dt=-e^{x}dx$:
$$\int\frac{du}{u}=-\int\frac{e^{x}}{1-e^{x}}dx
\quad\Longrightarrow\quad \ln|u|=\ln|1-e^{x}|+C.$$
Substitute back $u=\tan y$ and exponentiate:
$$\tan y=C(1-e^{x}),\qquad C\ \text{an arbitrary constant.}$$
Final Answer
$$\boxed{\tan y = C\big(1-e^{x}\big)}$$
Download clear revision notes from Anand Classes — perfect for JEE and CBSE practice, concise and reliable material for first-order differential equations.
NCERT Question.11 : Find a particular solution satisfying the given condition:
$$(x^{3}+x^{2}+x+1)\frac{dy}{dx}=2x^{2}+x, $$
$y = 1$ when $x = 0$
Solution :
Factor the denominator:
$$x^{3}+x^{2}+x+1=(x+1)(x^{2}+1).$$
So
$$\frac{dy}{dx}=\frac{2x^{2}+x}{(x+1)(x^{2}+1)}.$$
Perform partial fractions:
$$\frac{2x^{2}+x}{(x+1)(x^{2}+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^{2}+1}.$$
Multiplying through and comparing coefficients gives
$$2x^{2}+x=(A+B)x^{2}+(B+C)x+(A+C).$$
Equating coefficients:
$$
A+B=2, B+C=1, A+C=0
$$
Solving yields $A=\tfrac{1}{2},B=\tfrac{3}{2},C=-\tfrac{1}{2}$.
Thus
$$\frac{dy}{dx}=\frac{1/2}{x+1}+\frac{\tfrac{3}{2}x-\tfrac{1}{2}}{x^{2}+1}.$$
Separate and integrate:
$$y=\int\frac{1}{2}\frac{dx}{x+1}+\int\frac{\tfrac{3}{2}x}{x^{2}+1},dx+\int\frac{-\tfrac{1}{2}}{x^{2}+1},dx + C.$$
Evaluate the integrals:
$$
\int\frac{1}{2}\frac{dx}{x+1} = \tfrac{1}{2}\ln|x+1| $$
$$\int\frac{\tfrac{3}{2}x}{x^{2}+1}dx = \tfrac{3}{4}\ln\big(x^{2}+1\big) $$
$$\int\frac{-\tfrac{1}{2}}{x^{2}+1}dx = -\tfrac{1}{2}\tan^{-1}x.
$$
So the general solution is
$$y=\tfrac{1}{2}\ln|x+1|+\tfrac{3}{4}\ln\big(x^{2}+1\big)-\tfrac{1}{2}\tan^{-1}x + C.$$
Use the condition $y(0)=1$:
$$1=\tfrac{1}{2}\ln 1+\tfrac{3}{4}\ln 1-\tfrac{1}{2}\tan^{-1}0+C \Longrightarrow C=1.$$
Final (particular) solution
$$\boxed{y=\tfrac{1}{2}\ln(x+1)+\tfrac{3}{4}\ln\big(x^{2}+1\big)-\tfrac{1}{2}\tan^{-1}x+1}$$
Download concise revision notes by Anand Classes — perfect for JEE and CBSE practice and ideal for quick mastery of partial fractions and integration techniques.
NCERT Question.12 : Find a particular solution satisfying the given condition:
$$x(x^{2}-1)\frac{dy}{dx}=1,\qquad y(2)=0$$
Solution :
Rearrange:
$$\frac{dy}{dx}=\frac{1}{x(x^{2}-1)}=\frac{1}{x(x-1)(x+1)}.$$
Partial fraction decomposition: find $A,B,C$ with
$$\frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}.$$
Putting $x=0,1,-1$ gives
$$A=-1,\qquad B=\tfrac{1}{2},\qquad C=\tfrac{1}{2}.$$
Thus
$$\frac{dy}{dx}=-\frac{1}{x}+\frac{1}{2}\cdot\frac{1}{x-1}+\frac{1}{2}\cdot\frac{1}{x+1}.$$
Integrate:
$$
y=\int\left(-\frac{1}{x}+\frac{1}{2}\frac{1}{x-1}+\frac{1}{2}\frac{1}{x+1}\right)dx + C $$
$$y =-\ln|x|+\tfrac{1}{2}\ln|x-1|+\tfrac{1}{2}\ln|x+1|+C$$
$$y =\tfrac{1}{2}\ln\left|\frac{x^{2}-1}{x^{2}}\right|+C$$
Use $y(2)=0$:
$$0=\tfrac{1}{2}\ln\left(\frac{2^{2}-1}{2^{2}}\right)+C=\tfrac{1}{2}\ln\left(\frac{3}{4}\right)+C,$$
so
$$C=-\tfrac{1}{2}\ln\left(\frac{3}{4}\right)=\tfrac{1}{2}\ln\left(\frac{4}{3}\right).$$
Particular solution
$$\boxed{y=\tfrac{1}{2}\ln\left(\frac{x^{2}-1}{x^{2}}\right)+\tfrac{1}{2}\ln\left(\frac{4}{3}\right)
=\tfrac{1}{2}\ln\left(\frac{4(x^{2}-1)}{3x^{2}}\right)}$$
Download concise revision notes by Anand Classes — perfect for JEE and CBSE practice, a clear and compact review of partial fractions and first-order differential equations.
Summary
Chapter 9 of the Class 12 NCERT Mathematics Part II textbook, “Differential Equations,” focuses on techniques for solving differential equations. Exercise 9.3 provides practice problems on applying methods for solving both first-order and higher-order differential equations. Solutions are detailed to help students understand and effectively apply these techniques to various problems.
