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I have the following Cauchy problem:

$$ y' = \vert y \vert - \arctan{e^x} $$ $$ y(0) = y_0 $$

I want to prove that there exists a specific value of $y_0$ such that:

$$ \lim_{x\to\infty} y(x) = \frac{\pi}{2}.$$

I am not too sure how to even start on this, beside saying that f(x,y) is Lipschizt and therefore I have local uniqueness and existence of solution which can be extended globally to $\mathbb{R}$. Any help?

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  • $\begingroup$ As long as $y \geq 0$ you can study the linear ODE $y' = y - \arctan(e^x)$. $\endgroup$ – Christoph Jan 29 at 12:10
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Let $y($ be a solution and suppose that $y(x)=0$ for some $x\ge0$. Then $y'(x)<0$. This implies that once a solution takes a negative value, it remains negative, Thus, the only way to solve the problem is to look for positive solutions. This means solving the linear equation $y'=y-\arctan(e^x)$. Its solution is $$ y(x)=e^x\Bigl(y_0-\int_0^xe^{-t}\arctan(e^t)\,dt\Bigr). $$ This will be positive if and only if $$ y_0\ge \int_0^\infty e^{-t}\arctan(e^t)\,dt=a. $$ If $y_0>a$, then $\lim_{x\to\infty}y(x)=\infty$. If $y_0=a$, then $$ \lim_{x\to\infty}y(x)=\lim_{x\to\infty}e^x\int_x^\infty e^{-t}\arctan(e^t)\,dt=\frac\pi2. $$ Note. Although it is not needed for the argument, the actual value of $a$ is $$ \frac{\pi+\log4}{4}. $$

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  • $\begingroup$ If $y_0 = a $ wouldn't y(x) be always 0? $\endgroup$ – qcc101 Jan 31 at 10:04
  • $\begingroup$ No. If $x<\infty$, then $y_0>\int_0^xe^{-t}\arctan(e^t)\,dt$. $\endgroup$ – Julián Aguirre Jan 31 at 12:30

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