Is this initial value problem well-defined for every $T>0$?
$$y’(t)=-t^2y\cos(t)$$ $$y(0)=2$$
where $t\in[0,T]$.
I know that since $f(y,t)= -t^2y\cos(t)$ and its derivative $f_y(y,t)$ are continuous functions everywhere, there exists an interval $[0,T]$ in which the problem is well-defined. However, I’m not sure how to find the value of $T$.
Update: I found the solution $y(t) = 2 e^{-(t^2 - 2) \sin(t) - 2 t \cos(t)}$, but I’d like to know if there’s a way to obtain $T$ without previously calculating it.