Mathematics - Differential Equation Question with Solution | TestHub

The given equation can be written as $2y\sin x\frac{dy}{dx}+y^2\cos x=\sin 2x$ $\Rightarrow \frac{d}{dx}\left(y^2\sin x\right)=\sin 2x\Rightarrow y^2\sin x=(-1/2)\cos 2x+C$.

So $(y(\pi /2){)}^2\sin (\pi /2)=(-1/2)\cos (2\pi /2)+C\Rightarrow C=1/2$.

Hence $y^2\sin x=(1/2)(1-\cos 2x)={\sin }^{2}x\Rightarrow y^2=\sin x$