Linear First Order ODE/x y' - 3 y = x^4

Theorem
The linear first order ODE:
 * $(1): \quad x \dfrac {\d y} {\d x} - 3y = x^4$

has the solution:
 * $y = x^4 + \dfrac C {x^3}$

Proof
Rearranging $(1)$:
 * $(2): \quad \dfrac {\d y} {\d x} + \paren {-\dfrac 3 x} y = x^3$

$(2)$ is a linear first order ODE in the form:
 * $\dfrac {\d y} {\d x} + \map P x y = \map Q x$

where:
 * $\map P x = -\dfrac 3 x$
 * $\map Q x = x^3$

Thus:

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

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

or:
 * $y = x^4 + \dfrac C {x^3}$