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

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

has the general solution:
 * $-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$

Proof
Let $(1)$ be expressed as:


 * $(2): \quad \paren {y + x^2 y^3} \rd x + \paren {x - x^3 y^2} \rd y = 0$

We note that $(2)$ is in the form:
 * $\map M {x, y} \rd x + \map N {x, y} \rd y = 0$

but is not exact.

So, let:
 * $\map M {x, y} = y + x^2 y^3$
 * $\map N {x, y} = x - x^3 y^2$

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

Thus:

It can be observed that:

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

So Integrating Factor for First Order ODE: Function of Product of Variables can be used.

Let $z = x y$.

Then:
 * $\map \mu {x y} = \map \mu z = e^{\int \map g z \rd z}$

Hence:

Thus an integrating factor for $(1)$ has been found:
 * $\mu = \dfrac 1 {x^3 y^3}$

which yields, when multiplying it throughout $(2)$:
 * $\paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$

which is now exact.

By First Order ODE: $\paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$, its general solution is:
 * $-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$