Question

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

Answer 1

Given, differential equation is $\frac{dy}{dx}+ycotx=4xcosecx$ ...(i)
On comparing with the form $\frac{dy}{dx}+Py=Q,$ we get
P=cotx, Q=4x ~cosecx and $IF=e^{\int Pdx}$
$\therefore IF=e^{\int cotxdx}\Rightarrow IF=e^{log|sinx|}=sinx$
The general solution of the given differential equation is given by
$y.IF=\int Q\times IFdx+C\Rightarrow ysinx=\int 4xcosecxsinxdx+C\Rightarrow ysinx=\int 4xdx+C\Rightarrow ysinx=4.\frac{x^2}{2}+C\Rightarrow ysinx=2x^2+C$
Now, $x=\frac{\pi }{2}andy=0\therefore 0\times sin\frac{\pi }{2}=2{\left(\frac{\pi }{2}\right)}^2+C\Rightarrow C=-\frac{{\pi }^2}{2}$
On putting the value of C in Eq. (ii), we get
$ysinx=2x^2-\frac{{\pi }^2}{2}$