Answer
$y = x^3-\frac{C_1}{x}+C_2$
Work Step by Step
Let $u = y'$
A first-order linear differential equation is one that can be put into the form:
$\frac{dy}{dx} + P(x)~y = Q(x)$
We can consider the given equation:
$xy'' + 2y' = 12x^2$
$xu' + 2u = 12x^2$
$u' + \frac{2}{x}u = 12x$
Let $P(x) = \frac{2}{x}$
Let $Q(x) = 12x$
We can find the integrating factor:
$I(x) = e^{\int \frac{2}{x} ~dx} = e^{2ln~x} = e^{ln~x^2} = x^2$
We can multiply by the integrating factor:
$x^2~u' + 2x~u = 12x^3$
$\frac{d}{dx}(x^2~u) = 12x^3$
We can integrate both sides of the equation:
$x^2~u = \int 12x^3~dx$
$x^2~u = 3x^4+C_1$
$u = 3x^2+\frac{C_1}{x^2}$
We can substitute $u = y'$ and integrate both sides of the equation:
$u = 3x^2+\frac{C_1}{x^2}$
$y' = 3x^2+\frac{C_1}{x^2}$
$y = x^3-\frac{C_1}{x}+C_2$
