Answer
See below
Work Step by Step
Given: $\frac{dy}{dx}-\frac{3}{2x}y=6y^{\frac{1}{3}}x^2\ln x$
This is a Bernoulli equation, where $p(x)=-\frac{3}{2x}\\q(x)=6x^2\ln x\\n=\frac{1}{3}$
Divide both sides of the given equation by $y^{\frac{1}{3}}$, we have:
$^{-\frac{1}{3}}\frac{dy}{dx}-{\frac{3}{2x}}y^{\frac{2}{3}}=6x^2\ln x$
To solve this, we let $u=y^{\frac{2}{3}} \rightarrow \frac{du}{dx}=\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}$
the equation then becomes: $\frac{3}{2}\frac{du}{dx}-\frac{3}{2x}u=6x^2\ln x\\\frac{du}{dx}-\frac{1}{x}u=4x^2\ln x$
Integrating factor: $I(x)=e^{-\int \frac{1}{x}dx}=e^{-\ln|x|}=|x|^{-1}$
Hence, $\frac{d}{dx}(\frac{u}{x})=-\frac{1}{x^2}u+\frac{1}{x}\frac{du}{dx}\\
\rightarrow \frac{d}{dx}(\frac{u}{x})=4x\ln x$
Integrate: $\int \frac{d}{dx}(\frac{u}{x})=\int 4x\ln x$
Let $u=\ln x \rightarrow du=\frac{1}{x}dx\\dv=xdx \rightarrow v=\frac{x^2}{2}$
then $\frac{u}{x}=\int 4x \ln x\\
\rightarrow \frac{u}{x}=4(\frac{x^2}{2}\ln x-\int \frac{x^2}{2}\frac{1}{x}du)\\
\rightarrow \frac{u}{x}=2x^2\ln x-x^2\\
\rightarrow u=2x^3\ln x-x^3$
Substitute: $y^{\frac{2}{3}}=2x^3\ln x-x^3$
Hence, $y^{\frac{2}{3}}=x^3(2\ln x-1)+c$
with $c$ is a constant.
