Question

Find the particular solution of the differential equation $\frac{dy}{dx}-3y\cot x=\sin 2x$, given that y = 2 when $x=\frac{\pi }{2}$.

Answer 1

Given,

$\frac{dy}{dx}-3y\cot x=\sin 2x$

$\frac{dy}{dx}+(-3\cot x)y=\sin 2x$

$\frac{dy}{dx}+Py=Q$

$P=-3\cot x,Q=\sin 2x$

$I.F=e^{\int Pdx}$

$=e^{\int (-3\cot x)dx}$

$={\csc }^{3}x$

$y\times I.F=\int Q\times I.Fdx$

$y\times {\csc }^{3}x=\int \sin 2x{\csc }^{3}xdx$

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

$y{\csc }^{3}x=-2\csc x+c$

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

$y=2,x=\frac{\pi }{2}$

$2=-2{\sin }^2\left(\frac{\pi }{2}\right)+c{\sin }^3\left(\frac{\pi }{2}\right)$

$2=-2+c$

$\therefore c=4$

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