Question

Solution of the differential equation $2y\sin x\frac{dy}{dx}=$ $2\sin x\cos x-y^2\cos x$ satisfying $y\left(\frac{\pi }{2}\right)=1$ is given by

• A. $y^2=\sin x$
• B. $y={\sin }^{2}x$
• C. $y^2=\cos x+1$
• D. $y^2\sin x=4{\cos }^{2}x$

Answer 1

Answer: (A)

The given equation can be written as
$2y\sin x\frac{dy}{dx}+y^2\cos x=\sin 2x$
$\frac{d}{dx}\left(y^2\sin x\right)=\sin 2x$
On integrating, we get
$y^2\sin x=\frac{-1}{2}\cos 2x+c$
Put $x=\frac{\pi }{2},y=1$, we get $c=\frac{-1}{2}$
Hence, the solution is
$y^2\sin x=\frac{1}{2}(1-\cos 2x)={\sin }^{2}x$
$\Rightarrow y^2=\sin x$