Question

Let $y=y(x)$ 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 $y(\frac{\pi }{6})$ is equal to :

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

Answer 1

Answer: (C)

$\sin x\frac{dy}{dx}+y\cos x=4x,x\in (0,\pi )$
$\frac{dy}{dx}+y\cot x=\frac{4x}{\sin x}$
$\therefore$ I.F. $=e^{\int \cot xdx}=\sin x$
$\therefore$ Solution is given by
$y\sin x=\int \frac{4x}{\sin x}\cdot \sin xdx$
$y\cdot \sin x=2x^2+c$
when $x=\frac{\pi }{2},y=0$
$\Rightarrow c=-\frac{{\pi }^2}{2}$
$\therefore$ Equation is : $y\sin x=2x^2-\frac{{\pi }^2}{2}$
when $x=\frac{\pi }{6}$ then $y\cdot \frac{1}{2}=2\cdot \frac{{\pi }^2}{36}-\frac{{\pi }^2}{2}$
$\therefore y=-\frac{8{\pi }^2}{9}$