Question

If $y=y\left(x\right)$ is the solution of differential equation $y\sin x\frac{dy}{dx}=\cos x(\sin x-y^2)$ where $x\ne n\pi ,n\in I$ and $y\left(\frac{\pi }{2}\right)=\sqrt{\frac{2}{3}}.$ Then $9y^4\left(\frac{\pi }{3}\right)=$

Answer 1

Given : $y\sin x\frac{dy}{dx}=\cos x(\sin x-y^2)$
$\Rightarrow y\frac{dy}{dx}+y^2\cot x=\cos x$
put $y^2=z,2y\frac{dy}{dx}=\frac{dz}{dx}$
$\Rightarrow \frac{dz}{dx}+z\left(2\cot x\right)=2\cos x$ (linear form)
$I.F.=e^{\int 2\cot xdx}={\sin }^{2}x$
solution is given by,
$z\left({\sin }^{2}x\right)=\int (2\cos x\cdot {\sin }^{2}x)dx\Rightarrow y^2\left({\sin }^{2}x\right)=\frac{2{\sin }^{3}x}{3}+C$
As, $y\left(\frac{\pi }{2}\right)=\sqrt{\frac{2}{3}},C=0$
$\Rightarrow y^2=\frac{2\sin x}{3}\therefore 9y^4\left(\frac{\pi }{3}\right)=3$