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

Theorem
The linear first order ODE:
 * $(1): \quad \left({x^2 + y}\right) \mathrm d x = x \, \mathrm d y$

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

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

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

where:
 * $P \left({x}\right) = -\dfrac 1 x$
 * $Q \left({x}\right) = x$

Thus:

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

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

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