Consider the differential equation $y'=ty(3-y), t\ge0$, determine the initial value $y(t_0)=y_0$ so that the solution to the IVP is certain to exist. Consider the differential equation $y'=ty(3-y), t\ge0$, determine the initial value $y(t_0)=y_0$ so that a solution to the IVP is certain to exist. I’m wondering if there's a way to answer this without solving the differential equation?
 A: I'm assuming that you can answer the question for local solution and that the task is asking for global solutions on $\Bbb R_+$.
The roots $y=0$ and $y=3$ on the right side are constant solutions. Any solution starting between these values stays bounded by them and thus exists for all $t\ge0$.
A: Yes, there is one test which is as follows:
For the IVP $y'=f(t, y), y(t_0)=y_0$, the sufficient condition to have solutions is that $f(t, y) $ must be continuous in some rectangular domain containing $(t_0,y_0)$ and 
for the solution to be unique the sufficient condition is that the partial derivative $\frac{\partial (ty(1-y)) }{\partial y} $ must be continuous in some rectangular domain containing $(t_0,y_0)$ 
A: Using change of variable $s=t^2/2$, the equation is actually a logistic one
\begin{align*}
\dfrac{dy}{dt}&=ty(3-y)\\
\dfrac{dy}{d(t^2/2)}=\dfrac{dy}{tdt}&=y(3-y)\\
\dfrac{dy}{ds}&=y(3-y)\\
y&=\dfrac{3}{1+C e^{-3s}}\\[5mm]
y&=\dfrac{3}{1+C e^{-3t^2/2}}.
\end{align*}
Other solutions are $y=0$ and $y=3$.
Hope this helps!
