Linear First Order ODE/(2 y - x^3) dx = x dy

Theorem
The linear first order ODE:
 * $(1): \quad \paren {2 y - x^3} \rd x = x \rd y$

has the general solution:
 * $y = -x^3 + C x^2$

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

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

where $P \left({x}\right) = -\dfrac 2 x$.

Thus:

Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
 * $\map {\dfrac \d {\d x} } {\dfrac y {x^2} } = -1$

and the general solution is:
 * $\dfrac y {x^2} = -x + C$

or:
 * $y = -x^3 + C x^2$