Answer
$y=\sqrt \frac{\ln x}{3x^3 \ln x -x^3 +C}$
Work Step by Step
We are given:
$$2x(\ln x)y'-y=-9x^3y^3\ln x$$
Dividing both sides by $2x(\ln x)$
$$y'-\frac{1}{2x(\ln x)}y=-\frac{9}{2}x^2y^3$$
The equation is a Berrnoulli equation.
Dividing the last equation by $y^3$
$$y^{-3}\frac{dy}{dx}-\frac{1}{2x(\ln x)}y^{-2}=-\frac{9}{2}x^2$$
Assume that $u=y^{-2} \rightarrow \frac{du}{dx}=-2y^{-3}\frac{dy}{dx}$
The equation becomes:
$$-\frac{1}{2}\frac{du}{dx}-\frac{1}{2x(\ln x)}u=-\frac{9}{2}x^2$$
$$\frac{du}{dx}+\frac{1}{x)\ln x)}u=9x^2$$
Integrating factor:
$$I=e^{\int \frac{1}{x(\ln x)}dx}=e^{\int \ln x dx}=\ln x$$
The equation becomes:
$$\frac{d}{dx}(u\ln x)=9x^2 \ln x$$
Integrating both sides:
$$u \ln x=3x^3 \ln x -x^3 +C$$
$$u=\frac{3x^3 \ln x -x^3 +C}{\ln x}$$
$$y^{-2}=\frac{3x^3 \ln x -x^3 +C}{\ln x}$$
The general solution is:
$$y=\sqrt \frac{\ln x}{3x^3 \ln x -x^3 +C}$$
