Answer:
Option C
Explanation:
We have
$\sin x\frac{dy}{dx}+y\cos x=4x$
$\Rightarrow$ $\frac{dy}{dx}+y\cot x=4x\times cosec x$
This is a linear differential equation of form
$\frac{dy}{dx}+Py=Q$
where P= cot x, Q= 4x cosec x
Now, IF= $e^{\int_{}^{} Pdx}=e^{\cot x dx}=e^{\log_{}{} \sin x}=\sin x$
Solution of the differential equation is
$y. \sin x= \int_{}^{} 4x cosecx\sin xdx +C$
$\Rightarrow$ $y \sin x=\int_{}^{} 4xdx+C=2x^{2}+C$
Put $x=\frac{\pi}{2},y=0,$ we get
$C= \frac{-\pi^{2}}{2}$
$\Rightarrow$ $y \sin x=2x^{2}-\frac{\pi^{2}}{2}$
Put $x=\frac{\pi}{6}$
$y(\frac{1}{2})=2(\frac{\pi^{2}}{36})-\frac{\pi^{2}}{2}$
$\Rightarrow$ $y=\frac{\pi^{2}}{9}-\pi^{2}$
$\Rightarrow$ $y=-\frac{8\pi^{2}}{9}$
Alternative method
We have , $\sin x\frac{dy}{dx}+y\cos x=4x$ ,
which can be written as $\frac{d}{dx}(\sin x.y)=4x$
On integrating both sides, we get
$\int_{}^{}\frac{d}{dx}(\sin x.y)dx= \int_{}^{} 4x.dx$
$\Rightarrow$ $y.\sin x=\frac{4x^{2}}{2}+C$
$\Rightarrow$ $y.\sin x=2x^{2}+C$
Now, as y=0 when $x=\frac{\pi}{2}$
$\therefore$ C= $-\frac{\pi^{2}}{2}$
$\Rightarrow$ $y.\sin x=2x^{2}-\frac{\pi^{2}}{2}$
Now , putting $ x=\frac{\pi}{6}$, we get
$y(\frac{1}{2})=2(\frac{\pi^{2}}{36})-\frac{\pi^{2}}{2}$
$y= \frac{\pi^{2}}{9}-\pi^{2}=-\frac{8\pi^{2}}{9}$
