# How to you find the general solution of dy/dx=x^3-4x?

Nov 28, 2016

$y = \frac{1}{4} {x}^{4} - 2 {x}^{2} + C$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{3} - 4 x$

We can just separate the variables "as is":
$\int \mathrm{dy} = \int {x}^{3} - 4 x \mathrm{dx}$

And so integrating gives us;
$y = \frac{1}{4} {x}^{4} - 2 {x}^{2} + C$