Class 12 Maths NCERT Solutions Chapter 9 Ex 9.5 – Linear Differential Equations

Here P=2,\ Q=\sin x, so I.F. =e^{\int 2\,dx}=e^{2x}.

y\,e^{2x}=\displaystyle\int e^{2x}\sin x\,dx+C. Using the standard reduction formula, \displaystyle\int e^{2x}\sin x\,dx=\dfrac{e^{2x}(2\sin x-\cos x)}{5}.

I.F. =e^{\int 3\,dx}=e^{3x}.

y\,e^{3x}=\displaystyle\int e^{-2x}e^{3x}dx+C=\displaystyle\int e^x\,dx+C=e^x+C.

I.F. =e^{\int dx/x}=e^{\log x}=x.

y\cdot x=\displaystyle\int x^2\cdot x\,dx+C=\dfrac{x^4}{4}+C.

I.F. =e^{\int\sec x\,dx}=e^{\log|\sec x+\tan x|}=\sec x+\tan x.

y(\sec x+\tan x)=\displaystyle\int\tan x(\sec x+\tan x)dx+C=\displaystyle\int(\sec x\tan x+\tan^2x)dx+C.

=\displaystyle\int\sec x\tan x\,dx+\displaystyle\int(\sec^2x-1)dx+C=\sec x+\tan x-x+C.

Dividing by \cos^2x: \dfrac{dy}{dx}+y\sec^2x=\tan x\sec^2x, so I.F. =e^{\int\sec^2x\,dx}=e^{\tan x}.

y\,e^{\tan x}=\displaystyle\int\tan x\sec^2x\,e^{\tan x}dx+C. Substituting t=\tan x: \displaystyle\int t\,e^t\,dt=e^t(t-1).

So y\,e^{\tan x}=e^{\tan x}(\tan x-1)+C.

Dividing by x: \dfrac{dy}{dx}+\dfrac{2y}{x}=x\log x, so I.F. =e^{\int 2/x\,dx}=x^2.

y\,x^2=\displaystyle\int x^3\log x\,dx+C. By parts (u=\log x): \displaystyle\int x^3\log x\,dx=\dfrac{x^4}{4}\log x-\dfrac{x^4}{16}+C.

Dividing by x\log x: \dfrac{dy}{dx}+\dfrac{y}{x\log x}=\dfrac{2}{x^2}, so I.F. =e^{\int dx/(x\log x)}=e^{\log|\log x|}=\log x.

y\log x=\displaystyle\int\dfrac{2\log x}{x^2}dx+C. By parts (u=\log x): \displaystyle\int\dfrac{2\log x}{x^2}dx=-\dfrac{2\log x}{x}-\dfrac{2}{x}+C.

\dfrac{dy}{dx}+\dfrac{2xy}{1+x^2}=\dfrac{\cot x}{1+x^2}, so I.F. =e^{\int 2x/(1+x^2)dx}=e^{\log(1+x^2)}=1+x^2.

y(1+x^2)=\displaystyle\int\cot x\,dx+C=\log|\sin x|+C.

Dividing by x: \dfrac{dy}{dx}+y\left(\dfrac{1}{x}+\cot x\right)=1, so I.F. =e^{\int(1/x+\cot x)dx}=e^{\log x+\log\sin x}=x\sin x.

y\cdot x\sin x=\displaystyle\int x\sin x\,dx+C. Integrating by parts: \displaystyle\int x\sin x\,dx=\sin x-x\cos x+C.

This is linear in x rather than y. Rewriting: \dfrac{dx}{dy}=x+y, i.e. \dfrac{dx}{dy}-x=y, so I.F. =e^{\int -1\,dy}=e^{-y}.

x\,e^{-y}=\displaystyle\int y\,e^{-y}dy+C. By parts: \displaystyle\int y\,e^{-y}dy=-e^{-y}(y+1)+C.

Linear in x: \dfrac{dx}{dy}+\dfrac{x}{y}=y, so I.F. =e^{\int dy/y}=y.

x\,y=\displaystyle\int y\cdot y\,dy+C=\dfrac{y^3}{3}+C.

Linear in x: \dfrac{dx}{dy}=\dfrac{x}{y}+3y, i.e. \dfrac{dx}{dy}-\dfrac{x}{y}=3y, so I.F. =e^{-\int dy/y}=\dfrac{1}{y}.

\dfrac{x}{y}=\displaystyle\int 3y\cdot\dfrac{1}{y}dy+C=3y+C.

I.F. =e^{\int 2\tan x\,dx}=e^{2\log\sec x}=\sec^2x.

y\sec^2x=\displaystyle\int\sin x\sec^2x\,dx+C=\displaystyle\int\sec x\tan x\,dx+C=\sec x+C.

At x=\pi/3,y=0: \sec(\pi/3)=2, so 0=2+C\Rightarrow C=-2.

I.F. =e^{\int 2x/(1+x^2)dx}=1+x^2.

y(1+x^2)=\displaystyle\int\dfrac{dx}{1+x^2}+C=\tan^{-1}x+C.

At x=1,y=0: 0=\dfrac{\pi}{4}+C\Rightarrow C=-\dfrac{\pi}{4}.

I.F. =e^{-3\int\cot x\,dx}=e^{-3\log\sin x}=\csc^3x.

y\csc^3x=\displaystyle\int\sin 2x\csc^3x\,dx+C=\displaystyle\int\dfrac{2\cos x}{\sin^2x}dx+C=-\dfrac{2}{\sin x}+C.

At x=\pi/2,y=2: 2=-2+C\Rightarrow C=4. Multiplying through by \sin^3x:

The condition gives \dfrac{dy}{dx}=x+y, i.e. \dfrac{dy}{dx}-y=x, so I.F. =e^{-x}.

y\,e^{-x}=\displaystyle\int x\,e^{-x}dx+C=-e^{-x}(x+1)+C.

At (0,0): 0=-1+C\Rightarrow C=1. So y\,e^{-x}=-e^{-x}(x+1)+1.

The condition gives x+y=\dfrac{dy}{dx}+5, i.e. \dfrac{dy}{dx}-y=x-5, so I.F. =e^{-x}.

y\,e^{-x}=\displaystyle\int(x-5)e^{-x}dx+C=e^{-x}(4-x)+C.

At (0,2): 2=4+C\Rightarrow C=-2. So y\,e^{-x}=e^{-x}(4-x)-2.

Dividing by x: \dfrac{dy}{dx}-\dfrac{y}{x}=2x, so P=-\dfrac{1}{x}.

I.F. =e^{\int -1/x\,dx}=e^{-\log x}=\dfrac{1}{x}.

Dividing by 1-y^2: \dfrac{dx}{dy}+\dfrac{y}{1-y^2}x=\dfrac{ay}{1-y^2}, so P_1=\dfrac{y}{1-y^2}.

I.F. =e^{\int y/(1-y^2)\,dy}=e^{-\frac{1}{2}\log(1-y^2)}=(1-y^2)^{-1/2}.