First Order ODE/(exp x - 3 x^2 y^2) y' + y exp x = 2 x y^3

Theorem
The first order ordinary differential equation:


 * $(1): \quad \left({e^x - 3 x^2 y^2}\right) y' + y e^x = 2 x y^3$

is an exact differential equation with solution:


 * $y e^x - x^2 y^3 = C$

Proof
Let $(1)$ be expressed as:


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

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

Then:

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:
 * $f \left({x, y}\right) = C$

where:

Hence:

and:

Thus:
 * $f \left({x, y}\right) = y e^x + x^2 y^3$

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


 * $y e^x + x^2 y^3 = C$