Question

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

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

Answer 1

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

$\frac{d}{dx}\left(y\sin x\right)=4x$
$\Rightarrow y\sin x=4\frac{x^2}{2}+c$
$\Rightarrow y\sin x=2x^2+c$
Given $y\left(\pi /2\right)=0$
$c=-\frac{{\pi }^2}{2}$
Thus, $y\sin x=2x^2-\frac{{\pi }^2}{2}$
Now $y\left(\frac{\pi }{6}\right)=-\frac{8}{9}{\pi }^2$