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

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

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

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

$(2)$ is in the form:
 * $\dfrac {\mathrm d y}{\mathrm d x} + P \left({x}\right) y = Q \left({x}\right)$

where $P \left({x}\right) = -\dfrac 3 x$.

Thus:

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

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

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