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

Theorem
The first order ODE:
 * $(1): \quad \dfrac 1 {x^3 y^2} \rd x + \paren {\dfrac 1 {x^2 y^3} + 3 y} \rd y = 0$

is an exact differential equation with solution:
 * $-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$

Proof
Let $M$ and $N$ be defined as:


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


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

Then:

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

where:

Hence:

and:

Thus:
 * $\map f {x, y} = -\dfrac 2 {x^2 y^2} + \dfrac {3 y^2} 2$

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


 * $-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$