How to solve Bernoulli type equations?

Question

I'm looking for hints on how to solve this equation :

$$xy^2y'=x^2+y^3$$

I know it's a Bernoulli type differential equation, but I don't know how to deal with it.

Answer 1

I used variation of parameter to solve it

$$xy^2y'=x^2+y^3$$ $$\frac x3(y^3)'=x^2+y^3$$ $$\frac x3(y^3)'-y^3=x^2$$ Solving the homogeneous equation $$\frac x3(y^3)'=y^3$$ $$\int \frac {d(y^3)}{y^3}=\int \frac {3dx}x=3\ln |x|+K$$ $$\ln|y^3|=3\ln |x|+K$$ $$y^3=Kx^3$$ $$y^3=K(x)x^3 \implies \frac x3(K(x)x^3)'-K(x)x^3=x^2$$ $$\frac 13K'x^4=x^2$$ $$K'=\frac 3 {x^2} \implies K= -\frac 3x+C$$ $$\boxed{y^3=(Cx-3)x^2}$$

Answer 2

Try to substitute $u=y^3$, then $u'=3y^2y'$.

Or you can also try to do the $y=ux$ substitution.