Can someone help me with a differential equation?

dy/dx = y^2(4-y^2)

dy/dx = 4y^2-y^4

dy * ( 1 / (4y^2-y^4) ) = dx

integrate both sides. Left side is not simple. I looked it up on wolframs integrals.com

(-y*log(y-2) - y * log(y+2) + 4 ) / (16* y) = x

solving for y looks difficult.

There might be a way using the something similar to the method of undetermined coefficients, but that is used for solving a linear differential equation.

You can integrate 1/(y^2(4-y^2)) by partial fractions.

First factor the denomiator:

1/(y^2(2-y)(2+y)), then

solve for A, B C, and D:

(Ay+B)/y^2 + C/(2-y) + D/(2+y) = 1/(y^2(2-y)(2+y))

When you solve this, you get A=0, B=1/4, C=D=1/16.

This gives you three integrals that you can evaluate.

Are we trying to integrate 4y^2 - y^4?

I think it's

(4y^3) / 3 - (y^5) / 5 + C

maybe this will help...

[1/(4y^2 - y^4)]dy = dx

annnnnd integrate