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There is a differential equation given: $$x' = t^3 - x^3$$ This equation's solution can't be extended to the right.
We are to prove that with the given initial condition: $$x(t_0) = x_0$$ contains the interval $[t_0, \infty)$.
What's more the proof should be completed without solving the equation.
I tried to figure out how can it be proved but I failed.
I would appreciate any tips or hints.

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  • $\begingroup$ i wouldn't even know how to solve it :-) $\endgroup$ – gt6989b Mar 22 '18 at 21:36
  • $\begingroup$ @gt6989b Neither would I! :) But I wrote it just in case :) $\endgroup$ – Hendrra Mar 22 '18 at 21:37
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Proof by contradiction. Suppose that the solution does not extend to infinity to the right. Then consider the solution of the initial value problem with maximal interval of existence, say $x:]d,c[\to\mathbb{R}$, where $d$ can be $-\infty$ or real, but $c<\infty$ by assumption. Then it is known that $x(t)\to+\infty$ or $x(t)\to-\infty$ as $t\to c-$ (by a theorem on limiting behavior of solutions following from the existence and uniqueness theorem).
We treat the first case $x(t)\to+\infty$, the second is analogous. There is a $\delta>0$ such that $x(t)>|c|+1$ for $c-\delta<t<c$. We estimate $$t^3<c^3< |c|^3+1<x(t)^3\mbox{ for }c-\delta<t<c.$$ This implies that $x'(t)=t^3-x(t)^3<0$ on this interval. Hence $x(t)$ is decreasing on it. This contradict the fact that $x(t)\to+\infty$.

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The vector field, the direction of the derivative, pushes any solution towards the line $x=t$, as for $x>t$ the derivative is negative and for $x<t$ positive. This makes the solution bounded on bounded intervals, thus infinitely extensible.

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