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Q.
Solution of the differential equation $2 y \sin x \frac{d y}{d x}=$ $2 \sin x \cos x-y^{2} \cos x$ satisfying $y\left(\frac{\pi}{2}\right)=1$ is given by
Differential Equations
Solution:
The given equation can be written as
$2 y \sin x \frac{d y}{d x}+y^{2} \cos x=\sin 2 x$
$\frac{d}{d x}\left(y^{2} \sin x\right)=\sin 2 x$
On integrating, we get
$y^{2} \sin x=\frac{-1}{2} \cos 2 x+c$
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 2 x)=\sin ^{2} x $
$\Rightarrow y^{2}=\sin x$