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$y'=\dfrac{1}{x^2+y^2}$ where $y=f(x)$ and $x$ lies in $[1,\infty)$ and $f(1)=1$ and it is is differentiable in that interval

I don't know how to even proceed in this problem.

Even the range of $y$ is sufficient

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    $\begingroup$ A cheap observation: $f(x)$ is an increasing concave down function. $\endgroup$ – iamvegan Aug 21 '16 at 15:08
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    $\begingroup$ @iamvegan Since when vegan people are smart? $\endgroup$ – Von Neumann Aug 21 '16 at 15:47
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    $\begingroup$ @MathMan: Certainly you are not asked to solve the ODE. Read carefully the wording of the problem. It is not necessary to solve the ODE to answer. By the way, the analytical solving is possible but of height level: The $x(y)$ ODE is a Riccati equation involving Bessel functions in the solution. But you don't need it. $\endgroup$ – JJacquelin Aug 21 '16 at 15:52
  • $\begingroup$ @JJacquelin can i get its maximum value.Even that can be sufficient $\endgroup$ – MathMan Aug 21 '16 at 16:41
  • $\begingroup$ @iamvegan can u giveits maximum value? $\endgroup$ – MathMan Aug 21 '16 at 16:41
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What is wanted isn't clear. May be this :

$y'=\frac{1}{x^2+y^2}>0 $ implies that $y(x)$ is increasing.

$y(1)=1$ hence $y(x)>1$ . As a consequence :

$$y'=\frac{1}{x^2+y^2}<\frac{1}{x^2+1}$$

$$y(x)<y(1)+\int_1^x \frac{dt}{t^2+1}=1+\tan^{-1}(x)-\tan^{-1}(1)$$ $$y(x)<1+\frac{\pi}{2}-\frac{\pi}{4}=1+\frac{\pi}{4}$$

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  • $\begingroup$ Yes perhaps this is sufficient.actually it is. Thanks $\endgroup$ – MathMan Aug 22 '16 at 2:03
  • $\begingroup$ @MathMan The provided answer is not what the title suggests. I wouldn't mind if you would share the actual answer from your textbook $\endgroup$ – imranfat Aug 22 '16 at 21:13
  • $\begingroup$ @imranfat Well the answer what jacquelin gave .Even solving it is fine.But i am afraid i dont have the solution of this DE $\endgroup$ – MathMan Aug 23 '16 at 11:43
  • $\begingroup$ @Mathman : Nobody has the solution $y(x)$ of this ODE on explicit form. The analytical solving is possible but of height level : The related x(y) ODE is a Riccati equation involving Bessel functions in a complicated formula. In fact, with a formula involving Bessel functions, it is possible to express the solution on the inverse form $x(y)$ but not on the form $y(x)$. Obviously, the original wording of this problem was deliberately given with an ODE that students cannot solve, in order to oblige them to answer without having an explicit solution of the ODE. $\endgroup$ – JJacquelin Aug 23 '16 at 13:07

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