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Consider the Cauchy problem $y' = \frac{t+2}{t^2+y^2}$, $y(0)=1$, study the behavior of its solutions, if it possible, when $t\to +\infty$.

By Cauchy-Lipschitz there exists a unique local solution; moreover $f(t, y)=\frac{t+2}{t^2+y^2}$ is bounded on every set of the form $[t_1,t_2]\times\mathbb{R}$, when $t_2>t_1>0$ (by $M=\frac{t_2+2}{t_1^2}$), therefore the solution is defined around $+\infty$, let's say for $t>t_0$.
Let $\phi$ be the solution, $\phi$ is strictly increasing and we have $\lim_{t\to +\infty}\phi(t)/t=\lim_{t\to +\infty}\phi'(t)=0$, hence $\phi = o_{+\infty}(t)$.
I've also tried to study $\phi''$ for the convexity, but found nothing.
Is there anything more to say? Can we say if $\phi$ is bounded?
Thanks in advance

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Since $y$ is increasing, $y(t)\ge1$ and $$ y'(t)\le\frac{t+2}{t^2+1}\quad t\ge0. $$ Integrating we get $$ y(t)\le1+\frac12\,\log(t^2+1)+2\arctan t\le1+\pi+\log(t+1). $$ Then $$ y'(t)\ge\frac{t+2}{t^2+(\log(t+1)+1+\pi )^2}. $$ Integrating again, we see that $y$ is not bounded. Using $\log(t+1)\le t$, we see that in fact $y$ has logarithmic growth.

As for convexity, we have to decide on the sign of $$ t^2+y^2-(2\,t+2\,y\,y')(t+2)\le y^2-t^2-4\,t. $$ We have $y''(0)<0$, so that $y$ is concave for small $t$. On the other hand, $y''<0$ for $t$ large enough. I am sure that $y$ is concave, but I have not worked out the details.

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  • $\begingroup$ Can you please explain how $y(t) \geq 1$ and hence $y'(t)\le\frac{t+2}{t^2+1}\quad t\ge0.$? $\endgroup$ – BAYMAX Sep 7 '18 at 9:10
  • $\begingroup$ @BAYMAX $y$ is increasing and $y(0)=1$. $\endgroup$ – Julián Aguirre Sep 7 '18 at 9:26
  • $\begingroup$ Oh nice! now how can we think of $y$ is not bounded how easily you did that integration? $\endgroup$ – BAYMAX Sep 7 '18 at 9:34
  • $\begingroup$ I did not compute the integral. The right hand side is like $1/t$ at $\infty$, so the integral diverges. $\endgroup$ – Julián Aguirre Sep 7 '18 at 9:42
  • $\begingroup$ Great! why are we connecting concave / convex here? $\endgroup$ – BAYMAX Sep 7 '18 at 10:12
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The following is an extended comment to Julián Aguirre's answer, but it is too long, so I am putting it here.

We already know that $\phi \colon [0, \infty) \to \mathbb{R}$ and is strictly increasing, so it has a limit as $t \to \infty$ (perhaps that limit is infinity).

Suppose that the limit is finite, say $K$. We have $$ \phi'(t) \ge \frac{t}{t^2 + K^2} \quad \forall \, t \in [0, \infty) $$ (because $0 < \phi(t) < K$ for all $t \ge 0$). Consequently $$ \phi(t) = \phi(0) + \int\limits_{0}^{t} \phi'(s) \, ds \ge 1 + \int\limits_{0}^{t} \frac{s \, ds}{s^2 + K^2} = 1 - \ln{K} + \tfrac{1}{2} \ln(t^2 + K^2) \to \infty, $$ a contradiction.

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