WeBWorK Standalone Renderer

The differential equation ${y-3y^{5}} = \left({y^{3}+4x} \right)y'$ can be written in differential form: $M(x,y) \, dx + N(x,y) \, dy = 0$ where

$M(x,y)=$ , and
$N(x,y)=$ . (To receive credit both answers must be correct.)

The left-hand side $M(x,y) \, dx + N(x,y) \, dy$ becomes an exact differential if it is divided by $y^5$ . Integrate the resulting equation to obtain a solution of the differential equation:

$= C$ .

You can earn partial credit on this problem.