Answer
$y=\sqrt \frac{1}{(2\cos x+1) \sin^2x}$
Work Step by Step
Given:
$$y'+y \cot x=y^3\sin^3x$$
This equation is in form of Bernoulli equation, where $p(x)=\cot x, q(x)=\sin^3x$ and $n=3$
Dividing both sides by $y^{3}$
$$\frac{1}{y^3}\frac{dy}{dx}+\frac{1}{y^2}\cot x=\sin^3x$$
Let $u=y^{-2} \rightarrow \frac{du}{dx}=-2\frac{1}{y^3}\frac{dy}{dx}$
Then it becomes:
$$-\frac{1}{2}\frac{du}{dx}+u\cot x=\sin^3x$$
$$\frac{du}{dx}-2u \cot (x)=-2\sin^3x$$
The integrating factor is:
$$I(x)=e^{-2\int \cot xdx}=e^{-2\int \frac{\cos x}{\sin x}}$$
To solve this, we let:
$\sin x =t \rightarrow \cos x dx=dt$
Hence,
$I=e^{-2\int \frac{1}{t}dt}=e^{-2\ln t}=e^{-2\ln \sin x}=\frac{1}{\sin^2x}=\csc^2x$
The equation becomes:
$$\frac{d}{dx}(u\csc^2x)=(u\frac{1}{\sin^2 x})\frac{du}{dx}=u\frac{-2\cos x}{\sin^3x}+\frac{1}{\sin^2x}\frac{du}{dx}$$
$$\frac{d}{dx}(u\csc^2x)=-2\sin x$$
Integrating both sides:
$$\int u \csc^2x=2\cos x+C$$
where $C$ is a constant of integration
Subtitute when $u=y^{-2}$ we have:
$$\rightarrow y^{-2}=(2\cos x+C) \sin^2x$$
$$\rightarrow y=\sqrt \frac{1}{(2\cos x+C) \sin^2x}$$
We are given $y(\frac{\pi}{2})=1$
Substituting:
$$1=\frac{1}{(2\times0+C)(\frac{1}{2}[1-\cos\pi])}$$
$\rightarrow C=1$
Hence here, the solution is:
$$y=\sqrt \frac{1}{(2\cos x+1) \sin^2x}$$
