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Given: $y(x)$ is defined for $x \geq 1$ and satisfies $$y'=\frac{1}{x^2+y^2}, y(1)=1$$

Show that $$ y(x) < \frac{5\pi}{4} $$ for all $x \geq 1$

I don't see an easy way to solve for $y(x)$, and I don't know how to demonstrate that the function is always less than $5\pi/4$ without solving for $y(x)$. Any suggestions on how to approach this?

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2 Answers 2

up vote 1 down vote accepted

Since $y'(x)\geq 0$ then $y(x)\geq 1$ if $x\geq 1$. This implies that $$ y'\leq \frac{1}{x^2+1}, $$ or, by integrating, $$ y(x)\leq \tan^{-1}x-\frac \pi 4+1\leq 1+\frac \pi 4, $$ which is somewhat sharper than you are asking.

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you can find the expression of y, $y(x)-1=\int_{1}^{x}y^{'}dy=\int_{1}^{x}\frac{1}{x^2+y^2}dy=\frac{\pi/4-\arctan\frac{1}{x}}{x},x>=1$

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4  
Why does the last equality hold?, $x$ is not constant w.r.t $y$ –  Dennis Gulko Nov 5 '12 at 18:57

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