Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

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.

share|improve this answer
add comment

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

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.