Answer
$ y=\sqrt \frac{-2 \cos^4 x+C}{\sin x}$
Work Step by Step
Given:
$$2y'+ y \cot x=8y^{-1}\cos ^{3}x$$
$$2\frac{dy}{dx}+ y \cot x=8y^{-1}\cos ^{3}x$$
$$\frac{dy}{dx}+\frac{\cot x}{2}y=4y^{-1}\cos ^{3}x$$
This equation is in form of Bernoulli equation, where $p(x)=\frac{\cot x}{2}, q(x)=4y^{-1}\cos ^{3}x$ and $n=-1$
Multiplying both sides by $y$
$$y\frac{dy}{dx}+\frac{\cot x}{2}y^2=4\cos^3x$$
Let $u=y^{2} \rightarrow \frac{du}{dx}=2y\frac{dy}{dx}$
Then it becomes:
$$\frac{1}{2}\frac{du}{dx}+\frac{\cot x}{2}u=4 \cot^3 x$$
$$\frac{du}{dx}+\cot(x)u=8\cot^3x$$
The integrating factor is:
$$I(x)=e^{\int \cot x dx}=e^{\frac{\cos x}{\sin x}}=x$$
To solve this, we let:
$\sin x =t \rightarrow \cos x dx=dt$
Hence,
$I=e^{\int \frac{1}{t}dt}=e^{\ln u}=e^{\ln \sin x}=\sin x$
The equation becomes:
$$\frac{d}{dx}(u \sin x)=u\cos x + \sin x\frac{du}{dx}$$
$$\frac{d}{dx}(u \sin x)=8\sin x \cos ^3 x$$
Integrating both sides:
$$\int \frac{d}{dx}(u \sin x)=8 \int \sin x \cos ^3 xdx$$
For the left side, we let:
$v=\cos x \rightarrow dv=-\sin x dx$
then $$\int \frac{d}{dx}(u \sin x)=8-\int v^3 dv=8(\frac{-v^4}{4})=-2v^4=-2 \cos^4 x+C$$
where $C$ is a constant of integration
Then:
$$u\sin x=-2 \cos^4 x+C$$
$$u=\frac{-2 \cos^4 x+C}{\sin x}$$
Subtitute when $u=y^{2}$ we have:
$$\rightarrow y^{1-\pi}=\frac{-2 \cos^4 x+C}{\sin x}$$
$$\rightarrow y=\sqrt \frac{-2 \cos^4 x+C}{\sin x}$$
