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 $(1)$:
 * $x \dfrac {\mathrm d p} {\mathrm d x} - p = 3 x^2$

and divide through by $x$:
 * $\dfrac {\mathrm d p} {\mathrm d x} - \dfrac p x = 3 x$

From:
 * Linear First Order ODE: $y' - \dfrac y x = 3 x$

its solution is:
 * $p = 3 x^2 + C_1 x$

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

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