Finite Interval of Existence of Solution of ODE

Asked Feb 03 '23 at 06:17

Active Feb 03 '23 at 07:34

Viewed 60 times

I am struggling on how to prove that the interval of existence of a solution of an ODE is finite. For example, for $x'(t)=x^2+t^2$ where $x \in \mathbb{R}$, I have separated as follows $$\int\frac{1}{x^2+t^2}dx=\int 1 dt$$ $$\frac{1}{t}\arctan \left(\frac{x}{t} \right)=t+c$$ $$\arctan \left(\frac{x}{t} \right)=t^2+tc$$ so we get $$x(t)=t \tan(t^2+tc)$$ so $t^2+tc \in \left( \frac{-\pi}{2}, \frac{\pi}{2} \right)$ which is finite for any finite value of $c$ but I am unsure of whether I did it right or am even on the right track.

Thank you.

edited Feb 03 '23 at 07:00

Matthew Cassell

4,364
4
23
31

asked Feb 03 '23 at 06:17

carlosvera

I think that this regards to concrete problem. – monotone operator Feb 03 '23 at 06:29
You didn't separate variables properly. The integral with the measure $dx$ still has the variable $t$ in it. Let $x \to ty$ to get $$(ty)' = t^{2} (y^{2} + 1)$$ which is a Riccati ODE. – Matthew Cassell Feb 03 '23 at 06:56
@Matthew It is also Riccati in its original form, a quadratic polynomial in $y$ on the right side. – Lutz Lehmann Feb 03 '23 at 07:25
@LutzLehmann Yeah I know, I was only stating that the equation that I gave after the transformation was a Riccati ODE so the OP could find some information on it. – Matthew Cassell Feb 03 '23 at 08:18
@MatthewCassell : I wanted to refer to the original equation in $x$, $x'=x^2+t^2$ is in some sense the classical (non-trivial) example for a Riccati equation. – Lutz Lehmann Feb 03 '23 at 09:09

Finite Interval of Existence of Solution of ODE

0 Answers0