[Math] Determining the interval where the solution is valid

Question

I am given the initial value problem

$$
y' = \frac{1+3x^2}{3y^2-6y}
$$
given y(0)= 1

I have solved this and I got $y^3-3y^2 -x-x^3=-2$. How would I got about finding the interval in which the solution is valid? I know I should solve for y first but I'm having issues doing that as well.

Thank you!

Answer 1

We have $3y^2-6y\ne0$. Since $y(0)=1$, we know that $0<y<2$.

Put $y=0$ in $y^3-3y^2 -x-x^3=-2$, and we know $x=1$. Similarly, when $y=2$, we have $x=-1$.

Hence, $-1<x<1$ is the interval we want. However, it is not easy to solve for y. (See Wolfram Alpha.)