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

Theorem

$(1): \quad x \dfrac {\d y} {\d x} - 3y = x^4$

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

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:

$\ds \int \map P x \rd x$

$=$

$\ds \int -\frac 3 x \rd x$

$\ds $

$=$

$\ds -3 \ln x$

$\ds $

$=$

$\ds \ln x^{-3}$

$\ds \leadsto \ \ $

$\ds e^{\int P \rd x}$

$=$

$\ds e^{\ln x^{-3} }$

$\ds $

$=$

$\ds x^{-3}$

$\ds $

$=$

$\ds \frac 1 {x^3}$

Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:

$\map {\dfrac \d {\d x} } {\dfrac y {x^3} } = 1$

$\dfrac y {x^3} = x + C$

or:

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

$\blacksquare$