First Order ODE/(1 over x^3 y^2 + 1 over x) dx + (1 over x^2 y^3 - 1 over y) dy = 0

Theorem

The first order ordinary differential equation:

$(1): \quad \paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$

is an exact differential equation with solution:

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

Proof

Let:

$\map M {x, y} = \dfrac 1 {x^3 y^2} + \dfrac 1 x$

$\map N {x, y} = \dfrac 1 {x^2 y^3} - \dfrac 1 y$

Then:

\(\ds \dfrac {\partial M} {\partial y}\)

\(=\)

\(\ds -\dfrac 2 {x^3 y^3}\)

\(\ds \dfrac {\partial N} {\partial x}\)

\(=\)

\(\ds -\dfrac 2 {x^3 y^3}\)

Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.

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

$\map f {x, y} = C$

where:

\(\ds \dfrac {\partial f} {\partial x}\)

\(=\)

\(\ds \map M {x, y}\)

\(\ds \dfrac {\partial f} {\partial y}\)

\(=\)

\(\ds \map N {x, y}\)

Hence:

\(\ds f\)

\(=\)

\(\ds \map M {x, y} \rd x + \map g y\)

\(\ds \)

\(=\)

\(\ds \int \paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \map g y\)

\(\ds \)

\(=\)

\(\ds -\frac 1 {2 x^2 y^2} + \ln x + \map g y\)

and:

\(\ds f\)

\(=\)

\(\ds \int \map N {x, y} \rd y + \map h x\)

\(\ds \)

\(=\)

\(\ds \int \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y + \map h x\)

\(\ds \)

\(=\)

\(\ds -\frac 1 {2 x^2 y^2} - \ln y + \map h x\)

Thus:

$\map f {x, y} = -\dfrac 1 {2 x^2 y^2} + \ln x - \ln y$

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:

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

$\blacksquare$