Question

Solution of $2y\sin x\frac{dy}{dx}=2\sin x\cdot \cos x-y^2\cos x,x=\frac{\pi }{2},y=1$ is given by

• A. $y^2=\sin x$
• B. $y={\sin }^{2}x$
• C. $y^2=\cos x+1$
• D. None of these

Answer 1

The correct option is A $y^2=\sin x$

$2y\sin x\frac{dy}{dx}=2\sin x\cdot \cos x-y^2\cos x$
$\Rightarrow 2y\sin x\frac{dy}{dx}+y^2\cos x=2\sin x\cdot \cos x$
$\Rightarrow \sin x\frac{d}{dx}\left(y^2\right)+y^2\frac{d}{dx}\sin x=2\sin x\cdot \cos x$
$\Rightarrow \frac{d}{dx}\left(y^2\sin x\right)=2\sin x\cdot \cos x$
$\Rightarrow \int d\left(y^2\sin x\right)=\int \sin 2x\cdot dx$
$\Rightarrow y^2\sin x=-\frac{\cos 2x}{2}+c$
At $x=\frac{\pi }{2},y=1$
$1^2\cdot \sin \frac{\pi }{2}=-\frac{\cos \pi }{2}+c\Rightarrow c=\frac{1}{2}$
So, $y^2\sin x=-\frac{\cos 2x}{2}+\frac{1}{2}$
$\Rightarrow y^2=\frac{1-\cos 2x}{2\sin x}$
$\Rightarrow y^2=\sin x$