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.