I apologize for the problem too simple, but I'm not able at solving it. Consider the Cauchy problem $$ \left\{ \begin{array}{l} \dot x=x(t)^2+t\\ x(0)=0 \end{array} \right. $$ Show that its solution is not defined in $[0,3]$.

## 2 Answers

Assume that x(t) is a solution: then, as its derivative is non negative, $\forall t$, $x(t) \geq 0$, so $x(1) \geq 0= \tan 0$. Note moreover that $\forall t \geq 1$, $x(t) \geq x'(t)^2 +1$. This forces $x(t)$, for $t \geq 1$ to be bigger than $\tan (t-1)$, which is the solution of the problem $$y(1)=0 \ \ \ y'(t)=y(t)^2+1,$$ and so the solution cannot exist beyond $1+\frac{\pi}2 <3$.

This differential equation has a "closed-form" solution:

$$ x(t) = {\frac {\sqrt {3}{{\rm Ai}'\left(-t\right)}+{ {\rm Bi}'\left(-t\right)}}{\sqrt {3}{{\rm Ai}\left(-t\right)}+{ {\rm Bi}\left(-t\right)}}} $$ where Ai and Bi are Airy functions. Numerically, the denominator has a zero at $t \approx 1.986352707$, so the solution does not exist beyond that point.