# Show that

Question:

$\frac{d y}{d x}-3 y \cot x=\sin 2 x ; y=2$ when $x=\frac{\pi}{2}$

Solution:

The given differential equation is $\frac{d y}{d x}-3 y \cot x=\sin 2 x$.

This is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q($ where $p=-3 \cot x$ and $Q=\sin 2 x)$

Now, I.F $=e^{\int p d x}=e^{-3 \int \cot x d x}=e^{-3 \log |\sin x|}=e^{\log \left|\frac{1}{\mid \sin ^{3} x}\right|}=\frac{1}{\sin ^{3} x} .$

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

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

$\Rightarrow y \cdot \frac{1}{\sin ^{3} x}=\int\left[\sin 2 x \cdot \frac{1}{\sin ^{3} x}\right] d x+\mathrm{C}$'

$\Rightarrow y \operatorname{cosec}^{3} x=2 \int(\cot x \operatorname{cosec} x) d x+\mathrm{C}$

$\Rightarrow y \operatorname{cosec}^{3} x=2 \operatorname{cosec} x+\mathrm{C}$

$\Rightarrow y=-\frac{2}{\operatorname{cosec}^{2} x}+\frac{3}{\operatorname{cosec}^{3} x}$

$\Rightarrow y=-2 \sin ^{2} x+\mathrm{C} \sin ^{3} x$                       ...(1)

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

Therefore, we get:

$2=-2+C$

$\Rightarrow C=4$

Substituting C = 4 in equation (1), we get:

$y=-2 \sin ^{2} x+4 \sin ^{3} x$

$\Rightarrow y=4 \sin ^{3} x-2 \sin ^{2} x$

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