Question

$\begin{array}{c}\text{Let}y=y\left(x\right)\text{be the solution of the differential}\text{equation}\sin x\frac{dy}{dx}+y\cos x=4x,x\in (0,\pi )\\ \text{If}y\left(\frac{\pi }{2}\right)=0,\text{then}y\left(\frac{\pi }{6}\right)\text{is equal to}\end{array}$

• A. $\frac{4}{9\sqrt{3}}{\pi }^2$
• B. $\frac{-8}{9\sqrt{3}}{\pi }^2$
• C. $-\frac{8}{9}{\pi }^2$
• D. $-\frac{4}{9}{\pi }^2$

Answer 1

The correct option is C $-\frac{8}{9}{\pi }^2$

Given equation is,

$sinx\frac{dy}{dx}+ycosx=4x$

Divide above equation by sinx

$\to \frac{dy}{dx}+ycotx=\frac{4x}{sinx}$

P=ctx,Q=4xcosecx

I.F.=$e^{\int p.dx}$

=$e^{cotxdx}$

=$e^{log\left(sinx\right)}$

=sinx

Complete solution is

$y\times sinx=\int sinx\times 4xcosecxdx+c$

$y.sinx=\int 4xdx+c$

$ysinx=2x^2+c$$-\left(1\right)$

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

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

$c=\frac{-{\pi }^2}{2}$

$1\to ysinx=2x^2-\frac{{\pi }^2}{2}$

at $x=\frac{\pi }{6}$

$ysin\left(\frac{\pi }{6}\right)=2\frac{{\pi }^2}{36}-\frac{{\pi }^2}{2}$

$y=2(\frac{{\pi }^2}{18}-\frac{{\pi }^2}{2})$

$y=\frac{-8{\pi }^2}{9}$