I have this differential equation: y'=y+2y^3, I must resolve the cauchy problem with y(0)=1... is it so hard?

Question

How i can do it?

Answer 1

dy/(y(1+2y^2)=dx so
dy/y-2y/(1+2y^2)*dy= dx
lnIyI -lnI1+2y^2I*1/2 =x+C
At x=0
ln 1-1/2 *ln3 =C
C=-1/2ln3 and
lnIyI-1/2 lnI1+2y^2I= x-1/2ln3