First Order ODE/y dx + (x^2 y - x) dy = 0

Theorem
The first order ODE:
 * $(1): \quad y \, \mathrm d x + \left({x^2 y - x}\right) \mathrm d y = 0$

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

This can also be presented in the form:
 * $\dfrac {\mathrm d y} {\mathrm d x} + \dfrac y {x^2 y - x}$

Proof
We note that $(1)$ is in the form:
 * $M \left({x, y}\right) \mathrm d x + N \left({x, y}\right) \mathrm d y = 0$

but that $(1)$ is not exact.

So, let:
 * $M \left({x, y}\right) = y$
 * $N \left({x, y}\right) = x^2 y - x$

Let:
 * $P \left({x, y}\right) = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$

Thus:

It can be observed that:

Thus $\dfrac {P \left({x, y}\right)} {N \left({x, y}\right)}$ is a function of $x$ only.

So Integrating Factor for First Order ODE: Function of One Variable can be used:


 * $\mu \left({x}\right) = e^{\int g \left({x}\right) \mathrm d x}$

Hence:

Thus an integrating factor for $(1)$ has been found:
 * $\mu = \dfrac 1 {x^2}$

which yields, when multiplying it throughout $(1)$:
 * $\dfrac y {x^2} \, \mathrm d x + \left({y - \dfrac 1 x}\right) \, \mathrm d y = 0$

which is now exact.

Let $M$ and $N$ be redefined as:


 * $M \left({x, y}\right) = \dfrac y {x^2}$
 * $N \left({x, y}\right) = y - \dfrac 1 x$

By Solution to Exact Differential Equation, the solution to $(1)$ is:
 * $f \left({x, y}\right) = C$

where:

Hence:

and:

Thus:
 * $f \left({x, y}\right) = \dfrac {y^2} 2 - \dfrac y x$

and by Solution to Exact Differential Equation, the solution to $(1)$, after simplification, is:


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