I have the following cauchy problem:
$$ y(t)' = (y(t) - t^3 - 1)^3 $$ $$ y(0) = \alpha $$ Discuss the global existence when
$1)\,\, \alpha < 1 $
$2) \,\,\alpha = 1 $
I tried the following:
If you let be $F(y,t) = (y - t^3 - 1)^3$, the function $F(y,t)$ exist for every $(y,t) \in \Re \times \Re$, being continuous and then a lipschitz function. That means the equation has a unique solution given the initial conditions for every $[a,b] \subseteq \Re$ with $a < b$. I know the function isn't sublinear, but this is a sufficent condition. If I would be able of solve the problem I also would find the domain of the solution, but I can't solve it.
I had attemp making $z = y -t^3 - 1$, so I have
$$ z(t)' = z^3 - 3t^2 $$ $$ y(0) = \alpha - 1 $$
but I can't solve neither this. That's why I believe that I need to do this without solving it.