First Order ODE/(2 x y^3 + y cosine x) dx + (3 x^2 y^2 + sine x) dy

Theorem
The first order ordinary differential equation:


 * $(1): \quad 2 x y^3 + y \cos x + \left({3 x^2 y^2 + \sin x}\right) \dfrac {\mathrm d y} {\mathrm d x} = 0$

is an exact differential equation with solution:


 * $x^2 y^3 + y \sin x = C$

This can also be presented as:
 * $\dfrac {\mathrm d y} {\mathrm d x} = -\dfrac {2 x y^3 + y \cos x} {3 x^2 y^2 + \sin x}$

Proof
Let:
 * $M \left({x, y}\right) = 2 x y^3 + y \cos x$
 * $N \left({x, y}\right) = 3 x^2 y^2 + \sin x$

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

and by Solution to Exact Differential Equation, the solution to $(1)$ is:
 * $x^2 y^3 + y \sin x = C$