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)

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