Answer
$y=c_1\ln x +x^3+c^2$
Work Step by Step
We are given:
$\frac{d^2y}{dx^2}+\frac{1}{x}\frac{dy}{dx}=9x$
Let $t=\frac{dy}{dx}$
$\frac{dv}{dx}+\frac{1}{x}v=9x$
The general solution is:
$v=I^{-1}(c_1+ \int 9x I dx)$
Intergrating factor:
$I=e^{\int \frac{1}{x}dx}=e^{\ln x}=x$
Multiply both sides by the intergrating factor:
$\frac{dy}{dx}=x^{-1}(c_1+ \int 9x^2dx)=\frac{1}{x}c_1+3x^2$
Integrate both sides:
$y=c_1\ln x +x^3+c^2$
