Answer
\[y(x)=\frac{C+x^3(3\ln x-1)}{\ln x}\]
Work Step by Step
$y'+\frac{1}{x\ln x}y=9x^2$ ____(1)
$\frac{dy}{dx}+\frac{1}{x\ln x}y=9x^2$ ____(2)
This is linear differential equation
Integrating factor:-
$I(x)=e^{\int\frac{1}{x\ln x}dx}=e^{\ln(\ln x)}=\ln x$
Multiply (2) by $I(x)$
$\ln x\frac{dy}{dx}+\frac{1}{x}y=9x^2\ln x$
$\frac{d}{dx}[y\ln x]=9x^2\ln x$
Integrating,
$y\ln x=\int 9x^2\ln x\;dx+C$
$C$ is constant of integration
$y\ln x=9\left[\ln x.\frac{x^3}{3}-\int\frac{1}{x}.\frac{x^3}{3}dx\right]+C$
$y\ln x=9\left[\ln x.\frac{x^3}{3}-\frac{x^3}{9}\right]+C$
$y\ln x=x^3(3\ln x-1)+C$
$y=\frac{C+x^3(3\ln x-1)}{\ln x}$
Hence general solution of (1) is $y(x)=\frac{C+x^3(3\ln x-1)}{\ln x}$.
