Anand Classes provides a comprehensive, free-downloadable PDF of NCERT Solutions for Class 12 Maths Chapter 9 – Differential Equations, Exercise 9.5 (Set-1), covering the method of solving homogeneous first-order, first-degree differential equations with detailed step-by-step explanations to help students master this topic from the latest CBSE syllabus. Click the print button to download study material and notes.
NCERT Question.1 : Find the general solution of differential equation
$$\frac{dy}{dx} + 2y = \sin x$$
Solution :
The given differential equation is
$$\frac{dy}{dx} + 2y = \sin x$$
Standard form of linear differential equation: comparing with
$$\frac{dy}{dx} + Py = Q$$
we have $P = 2$ and $Q = \sin x$
Integrating factor (I.F.) is
$$\mu(x) = e^{\int Pdx} = e^{\int 2dx} = e^{2x}$$
Hence, the solution is
$$y\mu = \int \mu Qdx + C$$
or
$$y e^{2x} = \int e^{2x} \sin xdx + C \quad …(i)$$
Let
$$I = \int e^{2x} \sin xdx \quad …(ii)$$
Applying integration by parts
$$I = \frac{e^{2x}}{2} \sin x – \frac{1}{2} \int e^{2x} \cos xdx$$
Let
$$J = \int e^{2x} \cos xdx$$
Again applying integration by parts
$$J = \frac{e^{2x}}{2} \cos x + \frac{1}{2} \int e^{2x} \sin xdx = \frac{e^{2x}}{2} \cos x + \frac{1}{2} I$$
Substituting this in $I$
$$I = \frac{e^{2x}}{2} \sin x – \frac{1}{2} \left( \frac{e^{2x}}{2} \cos x + \frac{1}{2} I \right)$$
$$I = \frac{e^{2x}}{2} \sin x – \frac{e^{2x}}{4} \cos x – \frac{1}{4} I$$
$$\frac{5}{4} I = e^{2x} \left( \frac{1}{2} \sin x – \frac{1}{4} \cos x \right)$$
$$I = \frac{1}{5} e^{2x} (2 \sin x – \cos x)$$
Putting this value of $I$ in $(i)$
$$y e^{2x} = \frac{1}{5} e^{2x} (2 \sin x – \cos x) + C$$
Dividing by $e^{2x}$
$$y = \frac{1}{5} (2 \sin x – \cos x) + C e^{-2x}$$
Hence, the required general solution is
$$\boxed{y = \frac{1}{5} (2 \sin x – \cos x) + C e^{-2x}}$$
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NCERT Question.2 : Find the general solution of differential equation
$$\frac{dy}{dx} + 3y = e^{-2x}$$
Solution :
The given differential equation is
$$\frac{dy}{dx} + 3y = e^{-2x}$$
Standard form of linear differential equation comparing with
$$\frac{dy}{dx} + Py = Q$$
we have $P = 3$ and $Q = e^{-2x}$
Integrating factor (I.F.) is
$$\mu(x) = e^{\int Pdx} = e^{\int 3dx} = e^{3x}$$
Hence the solution is
$$y\mu = \int \mu Qdx + C$$
or
$$y e^{3x} = \int e^{3x} e^{-2x}dx + C$$
Simplify the integrand
$$\int e^{3x} e^{-2x}dx = \int e^{x}dx = e^{x}$$
Thus
$$y e^{3x} = e^{x} + C$$
Dividing by $e^{3x}$
$$y = e^{x} e^{-3x} + C e^{-3x}$$
$$y = e^{-2x} + C e^{-3x}$$
Hence the required general solution is
$$\boxed{y = e^{-2x} + C e^{-3x}}$$
NCERT Question.3 : Find the general solution of differential equation
$$\frac{dy}{dx} + \frac{y}{x} = x^2$$
Solution :
The given differential equation is
$$\frac{dy}{dx} + \frac{y}{x} = x^2$$
It is of the form
$$\frac{dy}{dx} + Py = Q$$
Comparing, we have $P = \dfrac{1}{x}$ and $Q = x^2$
Integrating factor (I.F.) is
$$\mu(x) = e^{\int Pdx} = e^{\int \dfrac{1}{x}dx} = e^{\log x} = x$$
The general solution is
$$y\mu = \int \mu Qdx + C$$
or
$$yx = \int x \cdot x^2dx + C$$
$$yx = \int x^3dx + C$$
$$yx = \frac{x^4}{4} + C$$
Dividing both sides by $x$
$$y = \frac{x^3}{4} + \frac{C}{x}$$
Hence, the required general solution is
$$\boxed{y = \frac{x^3}{4} + \frac{C}{x}}$$
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NCERT Question.4 : Find the general solution of differential equation
$$\frac{dy}{dx} + (\sec x)y = \tan x \qquad (0 \le x \le \pi/2)$$
Solution :
The given differential equation is
$$\frac{dy}{dx} + (\sec x)y = \tan x$$
It is of the form
$$\frac{dy}{dx} + P y = Q$$
Comparing, $P=\sec x$ and $Q=\tan x$
Integrating factor (I.F.) is
$$\mu(x)=e^{\int Pdx}=e^{\int \sec xdx}=e^{\log|\sec x+\tan x|}=\sec x+\tan x$$
The general solution is
$$y\mu=\int \mu Qdx + C$$
or
$$y(\sec x+\tan x)=\int (\sec x+\tan x)\tan xdx + C$$
Simplify the integrand
$$(\sec x+\tan x)\tan x=\sec x\tan x+\tan^2 x$$
Use $\tan^2 x=\sec^2 x-1$ so the integrand becomes
$$\sec x\tan x+\sec^2 x-1$$
Note $\dfrac{d}{dx}(\sec x+\tan x)=\sec x\tan x+\sec^2 x$ so the integrand is
$$\frac{d}{dx}(\sec x+\tan x)-1$$
Integrate
$$\int\left(\frac{d}{dx}(\sec x+\tan x)-1\right)dx=\sec x+\tan x – x$$
Thus
$$y(\sec x+\tan x)=\sec x+\tan x – x + C$$
Divide by $\sec x+\tan x$
$$y=1-\frac{x}{\sec x+\tan x}+\frac{C}{\sec x+\tan x}$$
Hence the required general solution is
$$\boxed{y=1-\frac{x}{\sec x+\tan x}+\frac{C}{\sec x+\tan x}}$$
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NCERT Question.5 : Find the general solution of differential equation
$$\cos^2 x\frac{dy}{dx} + y = \tan x \qquad (0 \le x \le \pi/2)$$
Solution :
The given differential equation is
$$\cos^2 x\frac{dy}{dx} + y = \tan x $$
Divide throughout by $\cos^2 x$ to make the coefficient of $\dfrac{dy}{dx}$ unity
$$\frac{dy}{dx} + y\sec^2 x = \sec^2 x\tan x$$
This is of the form
$$\frac{dy}{dx} + P y = Q$$
Comparing
$$P = \sec^2 x$$
$$Q = \sec^2 x\tan x$$
Integrating factor (I.F.) is
$$\mu(x)=e^{\int Pdx}=e^{\int \sec^2 xdx}=e^{\tan x}$$
The general solution is
$$y\mu = \int \mu Qdx + C$$
or
$$y e^{\tan x} = \int e^{\tan x}\sec^2 x\tan xdx + C \quad …(i)$$
Put $t=\tan x$ so that $dt=\sec^2 xdx$ and the integral becomes
$$ \int te^{t} dt$$
Integrate by parts
$$\int te^{t} dt = t e^{t} – \int e^{t}dt = t e^{t} – e^{t} + C = (t-1)e^{t} + C$$
Putting this in $(i)$ and reverting to $x$
$$y e^{\tan x} = (\tan x – 1)e^{\tan x} + C$$
Dividing by $e^{\tan x}$
$$\boxed{y = \tan x – 1 + C e^{-\tan x}}$$
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NCERT Question.6 : Find the general solution of differential equation
$$x\frac{dy}{dx} + 2y = x^2\log x$$
Solution :
The given differential equation is
$$x\frac{dy}{dx} + 2y = x^2\log x$$
Divide by $x$ to make the coefficient of $\dfrac{dy}{dx}$ unity
$$\frac{dy}{dx} + \frac{2}{x}y = x\log x$$
Compare with $\dfrac{dy}{dx} + Py = Q$ so $P=\dfrac{2}{x}$ and $Q=x\log x$
Integrating factor
$$\mu(x)=e^{\int Pdx}=e^{\int 2/xdx}=e^{2\log x}=x^2$$
Hence
$$y\mu=\int \mu Qdx + C$$
or
$$y x^2 = \int x^2\cdot x\log xdx + C = \int x^3\log xdx + C$$
Compute $\displaystyle\int x^3\log xdx$ by parts with $u=\log x$ and $dv=x^3dx$ so $du=\dfrac{1}{x}dx$ and $v=\dfrac{x^4}{4}$
$$\int x^3\log xdx = \frac{x^4}{4}\log x – \int \frac{x^4}{4}\cdot\frac{1}{x}dx = \frac{x^4}{4}\log x – \frac{1}{4}\int x^3dx$$
$$= \frac{x^4}{4}\log x – \frac{1}{4}\cdot\frac{x^4}{4} = \frac{x^4}{4}\log x – \frac{x^4}{16}$$
So
$$y x^2 = \frac{x^4}{4}\log x – \frac{x^4}{16} + C$$
Divide by $x^2$
$$y = \frac{x^2}{4}\log x – \frac{x^2}{16} + C x^{-2}$$
Equivalently
$$\boxed{y = \frac{x^2}{16}\bigl(4\log x – 1\bigr) + C x^{-2}}$$
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NCERT Question.7 : Find the general solution of differential equation
$$x\log x\frac{dy}{dx}+y=\frac{2}{x}\log x$$
Solution :
The given differential equation is
$$x\log x\frac{dy}{dx}+y=\frac{2}{x}\log x$$
Divide by $x\log x$ to make the coefficient of $\dfrac{dy}{dx}$ unity
$$\frac{dy}{dx}+\frac{1}{x\log x}y=\frac{2}{x^{2}}$$
Compare with $\dfrac{dy}{dx}+Py=Q$ so
$$P=\frac{1}{x\log x}\qquad Q=\frac{2}{x^{2}}$$
Integrating factor
$$\mu(x)=e^{\int Pdx}=e^{\int \frac{1}{x\log x}dx}=e^{\log(\log x)}=\log x$$
Multiply through by $\log x$ and integrate
$$y\log x=\int \log x\cdot\frac{2}{x^{2}}dx + C
=2\int \frac{\log x}{x^{2}}dx + C$$
Compute the integral by parts with $u=\log x$ and $dv=x^{-2}dx$ so $du=\dfrac{1}{x}dx$ and $v=-\dfrac{1}{x}$
$$\int \frac{\log x}{x^{2}}dx=-\frac{\log x}{x}-\frac{1}{x}+K$$
Thus
$$y\log x=2\left(-\frac{\log x}{x}-\frac{1}{x}\right)+C
=-\frac{2\log x}{x}-\frac{2}{x}+C$$
Divide by $\log x$
$$\boxed{y=-\frac{2}{x}-\frac{2}{x\log x}+\frac{C}{\log x}}$$
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NCERT Question.8 : Find the general solution of differential equation
$$(1+x^2)dy + 2xydx = \cot xdx$$
Solution :
The given differential equation is
$$(1+x^2)dy + 2xydx = \cot xdx$$
Divide by $dx$
$$(1+x^2)\frac{dy}{dx} + 2xy = \cot x$$
Divide by $1+x^2$ to make the coefficient of $\dfrac{dy}{dx}$ unity
$$\frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{\cot x}{1+x^2}$$
Compare with $\dfrac{dy}{dx} + P y = Q$ so
$$P = \frac{2x}{1+x^2}, \quad Q = \frac{\cot x}{1+x^2}$$
Integrating factor
$$\mu(x)=e^{\int Pdx}=e^{\int \dfrac{2x}{1+x^2}dx}=e^{\ln(1+x^2)}=1+x^2$$
Multiply through by $\mu$ and integrate
$$y(1+x^2)=\int (1+x^2)\frac{\cot x}{1+x^2}dx + C=\int \cot xdx + C$$
$$y(1+x^2)=\ln|\sin x| + C$$
Divide by $1+x^2$
$$\boxed{y=\frac{\ln|\sin x|}{1+x^2}+\frac{C}{1+x^2}}$$
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NCERT Question.9 : Find the general solution of differential equation
$$x\frac{dy}{dx}+y-x+xy\cot x=0\qquad(x\neq 0)$$
Solution :
Given differential equation is
$$x\frac{dy}{dx}+y-x+xy\cot x=0\qquad(x\neq 0)$$
Rewrite as
$$x\frac{dy}{dx}+y+xy\cot x=x$$
Divide by $x$ to make the coefficient of $\dfrac{dy}{dx}$ unity
$$\frac{dy}{dx}+\frac{1}{x}y+\cot xy=1$$
Compare with $\dfrac{dy}{dx}+Py=Q$ so
$$P=\frac{1}{x}+\cot x ,\qquad Q=1$$
Integrating factor
$$\mu(x)=e^{\int Pdx}=e^{\int\left(\frac{1}{x}+\cot x\right)dx}=e^{\ln x+\ln\sin x}=x\sin x$$
Solution using the integrating factor
$$y\mu=\int\mu Qdx+C$$
$$y(x\sin x)=\int x\sin xdx+C$$
Compute the integral by parts
$$\int x\sin xdx=-x\cos x+\sin x$$
Thus
$$y(x\sin x)=-x\cos x+\sin x+C$$
Divide by $x\sin x$
$$\boxed{y=-\cot x+\frac{1}{x}+\frac{C}{x\sin x}}$$
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