Question

The particular solution of the differential equation $\frac{dy}{dx}+y\cot x=2x+x^2\cot x,$ such that $y(\pi /2)=0$ is

• A. $y=\frac{{\pi }^2}{4\cos x},(x\ne 0)$
• B. $y=x^2-\frac{\pi }{2}\tan x$
• C. $y=\frac{2x}{\sin x}+\frac{1}{x^2},(x\ne 0)$
• D. $y=x^2-\frac{{\pi }^2}{4\sin x}(\sin x\ne 0)$

Answer 1

Answer: (D)

Given, $\frac{dy}{dx}+y\cot x=2x+x^2\cot x$
$IF=e^{\int \cot xdx}=e^{\log \sin x}=\sin x$
$\therefore$ Complete solution is $y.\sin x=\int \sin x(2x+x^2\cot x)dx+C$
$\Rightarrow$ $y.\sin x=2\int x\sin xdx+\int x^2\cos xdx+C$
$\Rightarrow$ $y.\sin x=2\int x\sin xdx+x^2\sin x$ $-2\int x\sin xdx+C$
$\Rightarrow$ $y\sin x=x^2\sin x+C$ ..(i)
$\because$ $y\left(\frac{\pi }{2}\right)=0$ i.e, at $x=\frac{\pi }{2},y=0$
Then from Eq. (i). $0.\sin \frac{\pi }{2}={\left(\frac{\pi }{2}\right)}^2\sin \frac{\pi }{2}+C$
$\Rightarrow$ $C=-\frac{{\pi }^2}{4}$
Putting the value of C in Eq. (i), we get
$y.\sin x=x^2\sin x-\frac{{\pi }^2}{4}.$
$\Rightarrow$ $y=x^2-\frac{{\pi }^2}{4\sin x}$
$(\sin x\ne 0)$