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I've been reading "A text book on ordinary differential equations" of Sahir Ahmad & Ambrossi. On page 37 we have the following excercise:

"Explain why $x'+\dfrac{\sin(t)}{e^t+1}x=0$ cannot has solution $x(t)$ such that $x(1)=1$ and $x(2)=-1$."

As far I know, $f(t,x)$ is continuos everywhere since $e^t\ne -1$ for every $t$, and $\frac{\partial f}{\partial x}$ doesn't depends on $x$, so its constant and therefore continous, then the existence and unicity theorem must apply on neighborhood of those points. Where am I wrong? Can anyone explain me? Im very stuck af this assignament.

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    $\begingroup$ As $\dfrac{\sin t}{e^t+1}$ remains positive, $x'$ and $x$ have opposite signs, and $x$ cannot decrease below zero. $\endgroup$ – Yves Daoust Mar 2 at 13:19
  • $\begingroup$ @YvesDaoust: Actually the sign of $\frac{\sin t}{e^t+1}$ does not matter. $\tilde x(t) = 0$ is a solution, therefore all other solutions are identically zero or have no zero at all. $\endgroup$ – Martin R Mar 3 at 18:09
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Let $F$ be an antiderivative of the function $ \frac{ \sin t}{e^t+1}.$ Then the general solution of the differential equation is given by

$$x(t)=C e^{-F(t)},$$

where $C \in \mathbb R.$

From $1=x(1)= Ce^{-F(1)}$ we get $C>0.$

But from $-1=x(2)=Ce^{-F(2)}$ we get $C<0.$

Contradiction !

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Hint:

$$\left|\frac{\sin(t)}{1+e^t}x-\frac{\sin(t)}{1+e^t}y\right|\leq |x-y|,$$ for all $x,y\in\mathbb R$.

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Yes, the conclusion follows directly from the uniqueness part of the Picard–Lindelöf theorem for initial value problems.

  • The differential equation is of the form $ x'(t) = f(t, x(t))$ where $f(t, x) = -\frac{\sin(t)}{e^t+1} x$ is continuous, and uniformly Lipschitz continuous in $x$.
  • $\tilde x(t) = 0$ is a solution of that differential equation on the interval $I = [1, 2]$.

If $x$ is any solution of the differential equation on $I$ with $x(t_0) = 0$ then $x$ and $\tilde x$ are both solutions of the same initial value problem $x'(t) = f(t, x(t))$, $x(t_0) = 0$, and therefore $x = \tilde x$.

It follows that any solution of the differential equation on $I$ is either identically zero, or has no zeros at all.

In particular, there can be no solution which takes both positive and negative values.

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