# How to find out where a solution to a differential equation is defined?

I have a question concerning 1st order ODE's:

Let me assume that we have an ODE $y'(t)=g(y(t))+h(t)$.

And now I am asked to solve it and then determine where the solution is defined. Actually I do not know whether the question "where a solution is defined" is just a property of the parameter or also of the initial condition?- I guess both?

Now I was wondering whether the last thing is just a property of the solution or also of the process of solving this ODE. So what I am actually asking is: How would one do this? My idea was to write down each restriction to the solution that could occur on the way to the solution of the ODE. So if I had to divide by y somewhere but the solution would be totally okay with y=0, I would still have to cross it out.

• Have you read the Picard–Lindelöf theorem? – Pragabhava Jul 8 '13 at 16:45
• what are you trying to tell me? – user66906 Jul 8 '13 at 16:59
• The answer to your question is that theorem. Any ODE book will have a detailed proof, along with the relation between existence, uniqueness and continuation of solutions depending on parameters and initial conditions. – Pragabhava Jul 8 '13 at 17:04

Yes, of both. A simple example: $y'=cy^2$, $y(0)=y_0$. Assuming $c>0$ and $y_0>0$, the solution is $$y=\frac{1}{y_0^{-1}-ct}$$ which ceases to exist (blows up) at $y=1/(cy_0)$. The blow-up time depends on both $c$, a parameter in the equation, and the initial condition $y_0$. Informally: the faster it grows and the bigger it is to begin with, the faster it will blow up.
As solution is unique (which it usually is, by Picard–Lindelöf), the blow-up times, both forward and backward in time, are uniquely determined by the initial value problem. They do not depend on how we solve the equation. The maximal interval of existence of solution is located between two blow-up times: $(T_b,T_f)$. In my example it is $(-\infty, 1/(cy_0))$, since there is no blow-up going backward in time.