# Find a particular solution of the differential equation

Question:

Find a particular solution of the differential equation $\frac{d y}{d x}+y \cot x=4 x \operatorname{cosec} x(x \neq 0)$, given that $y=0$ when $x=\frac{\pi}{2}$

Solution:

The given differential equation is:

$\frac{d y}{d x}+y \cot x=4 x \operatorname{cosec} x$

This equation is a linear differential equation of the form

$\frac{d y}{d x}+p y=Q$, where $p=\cot x$ and $Q=4 x \operatorname{cosec} x$.

Now, I.F $=e^{\int \rho d x}=e^{\int \cot x d x}=e^{\log |\sin x|}=\sin x$

The general solution of the given differential equation is given by,

$y(\mathrm{IF} .)=\int(\mathrm{Q} \times \mathrm{I.F}) d x+\mathrm{C}$

$\Rightarrow y \sin x=\int(4 x \operatorname{cosec} x \cdot \sin x) d x+\mathrm{C}$

$\Rightarrow y \sin x=4 \int x d x+\mathrm{C}$

$\Rightarrow y \sin x=4 \cdot \frac{x^{2}}{2}+\mathrm{C}$

$\Rightarrow y \sin x=2 x^{2}+\mathrm{C}$             $\ldots(1)$

Now, $y=0$ at $x=\frac{\pi}{2}$

Therefore, equation (1) becomes:

$0=2 \times \frac{\pi^{2}}{4}+\mathrm{C}$

$\Rightarrow \mathrm{C}=-\frac{\pi^{2}}{2}$

Substituting $\mathrm{C}=-\frac{\pi^{2}}{2}$ in equation (1), we get:

$y \sin x=2 x^{2}-\frac{\pi^{2}}{2}$

This is the required particular solution of the given differential equation.