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This is part of an old exam:

Given the following differential equation: \begin{align} \dot{x} = -e^x \cdot \tan(x), \hspace{1cm} x(t_0) = x_0 \end{align}

Prove that every solution $x(t)$ with $x(0) = -1$ is strictly monotonously increasing.

Already proven are the following statements:

For each $(t_0, x_0) \in \mathbb{R} \times \left]-\frac{\pi}{2}, \frac{\pi}{2}\right[$ there exists a unique solution and $x(t) = 0$ is the solution fulfilling $x(0) = 0$.

Has anyone tips on how to solve this? Can the already proven facts somehow be used in order to show the statements?

Edit: Figures in order to get a feeling for the solution have been drawn. I did not succeed using Separation of variables to compute $x(t)$. If I do not miss something obvious I think it is hard to compute $x(t)$.

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    $\begingroup$ The question is probably meant to be solved without computing $x(t)$ explicitly. You have to prove positiveness of the derivative. $\endgroup$
    – user65203
    May 22 '19 at 12:33
  • $\begingroup$ Yeah, I think so as well after I failed to compute it. Nevertheless I have no idea how to solve it ... $\endgroup$
    – offline
    May 22 '19 at 12:35
  • $\begingroup$ Try to use my hint. $\endgroup$
    – user65203
    May 22 '19 at 12:37
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    $\begingroup$ Since $-e^x\tan x > 0$ for $x<0$, you have to prove $x(t)$ remains negative for all $t$. Maybe show that $x(t_0) = 0$ is impossible for your solution. $\endgroup$
    – GEdgar
    May 22 '19 at 12:43
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    $\begingroup$ Indeed by unicity of solutions to diff eq's, $x=0$ identically is the only solution to your diff eq and $x(s)=0$ for some $s$ (@GEdgar's hint). But - is the "strictly increasing" only for $t\ge 0$? Am I missing something? is there any reason for $x(u) =-\pi/2$ not happening for some $u<0$? $\endgroup$
    – peter a g
    May 22 '19 at 13:04
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An ODE of type $$\dot x=f(x)\qquad (t,x)\in{\mathbb R}\times\>]a,b[$$ with continuous $f$ has special solutions coming from the zeros $c_k$ $(1\leq k\leq m)$ of $f$. These zeros lead to the $m$ constant solutions $x(t)\equiv c_k$. The graphs of these solutions partition the rectangle ${\mathbb R}\times\>]a,b[$ into horizontal strips. Any other solution stays in a particular of these strips since its graph cannot cross a graph $x(t)\equiv c_k$. In particular any such solution is either monotonically increasing or monotonically decreasing since ${\rm sgn} f(x)$ is constant in each of these strips. For the initial point of your problem this sign is positive. Therefore the solution is strictly increasing as long as it lives. – Similarly, the solution through the initial point $(t_0,x_0):=(0,1)$ would be monotonically decreasing.

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  • $\begingroup$ That is a great answer! Thank you! Just to make sure: If it would be $x(0) = 1$ instead of $x(0) = -1$ the solution would be strictely monotonously decreasing, right? $\endgroup$
    – offline
    May 22 '19 at 15:32
  • $\begingroup$ As "peter a g" notes, in this case $f(t)$ also changes sign when $t=-\pi/2$. Numerically, it seems this happens at $t \approx -0.053$. Our solution has a vertical tangent there, but can it match another solution coming from $t<-0.053$? $\endgroup$
    – GEdgar
    May 22 '19 at 15:41

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