Linear First Order ODE/y' - (y over x) = 3 x

Theorem
The linear first order ODE:
 * $(1): \quad \dfrac {\mathrm d y} {\mathrm d x} - \dfrac y x = 3 x$

has the solution:
 * $\dfrac y x = 3 x + C$

or:
 * $y = 3 x^2 + C x$

Proof
$(1)$ is in the form:
 * $\dfrac {\mathrm d y}{\mathrm d x} + P \left({x}\right) y = Q \left({x}\right)$

where $P \left({x}\right) = -\dfrac 1 x$.

Thus:

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

and the general solution is:
 * $\dfrac y x = 3 x + C$

or:
 * $y = 3 x^2 + C x$