Linear First Order ODE/(x^2 + y) dx = x dy

Theorem
The linear first order ODE:
 * $(1): \quad \paren {x^2 + y} \rd x = x \rd y$

has the general solution:
 * $y = x^2 + C x$

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

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

where:
 * $\map P x = -\dfrac 1 x$
 * $\map Q x = x$

Thus:

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

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

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