Question

Solve the differential equation
$\frac{dy}{dx}-3y\cot x=\sin 2x$, given $y=2$, when $x=\frac{\pi }{2}$.

Answer 1

$\frac{dy}{dx}-3y\cot x=\sin 2x$
This is a linear equation of the form $\frac{dy}{dx}+Py=Q$
Where $P=-3\cot x,Q=\sin 2x$
$\therefore I.F.=e^{\int pdx}$
$I.F.=e^{\int -3\cot xdx}=e^{-3\int \cot xdx}$
$=e^{-3\log \sin x}=e^{\log (\sin x{)}^{-3}}$
$=(\sin x{)}^{-3}=\frac{1}{{\sin }^{3}x}$
$\therefore$ Its solution is $y.(I.F.)=\int Q(I.F.)dx$

$y.\frac{a}{{\sin }^{3}x}=\int \sin 2x.\frac{1}{{\sin }^{3}x}dx$

$\frac{y}{{\sin }^{3}x}=\int \frac{2\sin x.\cos x}{{\sin }^{3}x}dx$

$\frac{y}{{\sin }^{3}x}=2\int \text{cosec}x.\cot x.dx$

$\frac{y}{{\sin }^{3}x}=-2\text{cosec}x+c$

Putting $y=2,x=\pi /2,$ $\frac{2}{{\sin }^3\frac{\pi }{2}}=-2\text{cosec}\left(\frac{\pi }{2}\right)+c$

$2=-2+c\Rightarrow c=4$
$\therefore \frac{y}{{\sin }^{3}x}=-\frac{2}{\sin x}+4$
$y=-2{\sin }^{2}x+4{\sin }^{3}x$