I'm reading this notes about the continuity of solutions with respect to initial conditions and parameters: http://www.math.pitt.edu/~sph/1275/1275notes7-13.pdf.
Here, they say:
Here is another example:
$$\begin{cases}y'=ky^2 \\y(0)=\alpha\end{cases}$$
Using separation of variables we can show that the unique solution is
$$\phi(t,\alpha,k)=\frac{\alpha}{1-\alpha kt}.$$
If $\alpha=0$, then $\phi=0$ on $-\infty<t<\infty.$ If $\alpha k>0,$ then the solution exists on $(-\infty,1/\alpha k)$.If $\alpha k<0,$ then the solution exists on $(1/\alpha k,\infty)$.
The last part is the part that I don0t understand. Why the solution if $\alpha k>0$ is defined on $(-\infty,1/\alpha k)$, and not on $(1/\alpha k,\infty)$?
Thanks.
When $k,\alpha$ have the same sign, bad things happen going forward in time: $|y|$ grows faster and faster until finite time blowup. Going backward in time, $y$ just goes to zero and so nothing bad happens.
When they have the opposite sign, everything is reversed.
Another way to see this is just to note that you have a formula for the exact solution in hand. In general the maximal domain of the solution is the largest interval contained in the domain of the formula for the solution which contains the initial time (i.e. $0$). This idea that the solution and the formula for the solution can have different domains is often rather subtle for students, so you should think carefully about it.