Question

Find a particular solution of the differential equation $\frac{dy}{dx}+ycotx=4x\,cosec\, x(x≠0),$ given that $y=0$ when $x=\frac{π}{2}$

Answer 1

The correct answer is: $ysinx=2x^2-\frac{π^2}{2}$
The given differential equation is:
$\frac{dy}{dx}+ycotx=4x\,cosec\,x$
This equation is a linear differential equation of the form
$\frac{dy}{dx}+py=Q$,where $p=cotx$ and $Q=4x\, cosecx.$
Now,$I.F.=e^{∫pdx}=e^{∫cot xdx}=e^{log|sinx|}=sinx$
The given solution of the given differential equation is given by,
$y(I.F.)=∫(Q×I.F.)dx+C$
$⇒ysinx=∫(4x\,cosecx.sinx)dx+C$
$⇒ysinx=4∫xdx+C$
$⇒ysinx=4.\frac{x^2}{2}+C$
$⇒ysinx=2x^2+C...(1)$
Now,$y=0$ at $x=\frac{π}{2}$
Therefore,equation(1)becomes:
$0=2×\frac{π^2}{4}+C$
$⇒C=\frac{-π^2}{2}$
Substituting $C=\frac{-π^2}{2}$ in equation(1),we get:
$ysinx=2x^2-\frac{π^2}{2}$
This is the required particular solution of the given differential equation.