Exact differential equation

Problem:

$3x^2y dx + x^3dy = 0$

Solution:

We have $\frac{\partial 3x^2y}{\partial y} = 3x^2 = \frac{\partial x^3}{\partial x}$, therefore, we can find $f(x, y)$ such that $\frac{\partial f}{\partial x} = 3x^2y$ and $\frac{\partial f}{\partial y} = x^3$

It is quite obvious that $f(x, y) = x^3y + c$.

The answer to the differential equation is therefore $y = cx^{-3}$.

As a quick check, we have $\frac{dy}{dx} = -3cx^{-4}$, therefore

$3x^2y dx + x^3dy = 3x^2(cx^{-3})dx + x^3(-3cx^{-4})dx = 0$.