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:


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

is an exact differential equation with solution:


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

Proof
Let $(1)$ be expressed as:


 * $\paren {y e^x - 2 x y^3} \rd x + \paren {e^x - 3 x^2 y^2} \rd y = 0$

Let:
 * $\map M {x, y} = y e^x - 2 x y^3$
 * $\map N {x, y} = e^x - 3 x^2 y^2$

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

where:

Hence:

and:

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

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


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