First Order ODE/y dx + (x^2 y - x) dy = 0/Proof 1

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

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

but that $(1)$ is not exact.

So, let:
 * $\map M {x, y} = y$
 * $\map N {x, y} = x^2 y - x$

Let:
 * $\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$

Thus:

It can be observed that:

Thus $\dfrac {\map P {x, y} } {\map N {x, y} }$ is a function of $x$ only.

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


 * $\map \mu x = e^{\int \map g x \rd x}$

Hence:

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

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

which is now exact.

By First Order ODE: $\dfrac y {x^2} \rd x + \paren {y - \dfrac 1 x} \rd y = 0$, its solution is:
 * $\dfrac {y^2} 2 - \dfrac y x = C$