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

Theorem
The first order ODE:
 * $(1): \quad \left({x^2 y^3 + y}\right) \mathrm d x = \left({x^3 y^2 - x}\right) \mathrm d y$

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

Proof
Let $(1)$ be expressed as:


 * $(2): \quad \left({y + x^2 y^3}\right) \mathrm d x + \left({x - x^3 y^2}\right) \mathrm d y = 0$

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

but is not exact.

So, let:
 * $M \left({x, y}\right) = y + x^2 y^3$
 * $N \left({x, y}\right) = x - x^3 y^2$

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

Thus:

It can be observed that:

Thus $\dfrac {P \left({x, y}\right)} {M \left({x, y}\right)}$ 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:
 * $\mu \left({x y}\right) = \mu \left({z}\right) = e^{\int g \left({z}\right) \mathrm d z}$

Hence:

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

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

which is now exact.

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


 * $M \left({x, y}\right) = \dfrac 1 {x^3 y^2} + \dfrac 1 x$
 * $N \left({x, y}\right) = \dfrac 1 {x^2 y^3} - \dfrac 1 y$

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

where:

Hence:

and:

Thus:
 * $f \left({x, y}\right) = -\dfrac 1 {2 x^2 y^2} + \ln x - \ln y$

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


 * $-\dfrac 1 {2 x^2 y^2} + \ln x - \ln y = C$

or:
 * $-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$