Second Order ODE/x y'' - y' = 3 x^2

Theorem
The second order ODE:
 * $(1): \quad x y'' - y' = 3 x^2$

has the solution:
 * $y = x^3 + \dfrac {C_1 x^2} 2 + C^2$

Proof
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable.

Substitute $p$ for $y'$ in $(1)$:
 * $(2): \quad x \dfrac {\mathrm d p} {\mathrm d x} - p = 3 x^2$

Let $(2)$ be rearranged as:
 * $(3): \quad \dfrac {\mathrm d p} {\mathrm d x} - \dfrac 1 p x = 3 x$

It can be seen that $(2)$ is a linear first order ODE in the form:
 * $\dfrac {\mathrm d p}{\mathrm d x} + P \left({x}\right) x = Q \left({x}\right)$

where:
 * $P \left({x}\right) = -\dfrac 1 x$
 * $Q \left({x}\right) = 3 x$

Thus:

Thus from Solution by Integrating Factor, $(2)$ can be rewritten as:
 * $\dfrac {\mathrm d} {\mathrm d x} \left({\dfrac p x}\right) = \dfrac 1 x 3 x = 3$

and thence:
 * $\dfrac p x = 3 x + C_1$

Substituting back for $p$:
 * $\dfrac {\mathrm d y}{\mathrm d x} = 3 x^2 + C_1 x$

which is separable, leading to:
 * $y = x^3 + \dfrac {C_1 x^2} 2 + C^2$