# Differential equation! !

1. Dec 5, 2006

### singedang2

differential equation! urgent!

1. The problem statement, all variables and given/known data
we have $$f'(x) = \frac{1}{x^2 + f(x)^2}$$
and i need to show that $$|f(x)| \leq \frac{5 \pi}{4}$$ when $$x \geq 1$$

2. Relevant equations

3. The attempt at a solution

i know this has something to do with arctan, (by the looks of f'(x), it looks similar to arctan derivative) but i really don't know how to attempt this. help pls thanks!

Last edited: Dec 5, 2006
2. Dec 5, 2006

### chaoseverlasting

I dont know how to get a trigonometric funtion in this but if y=f(x), then $$dy/dx=f'(x)$$
This implies that $$y^2dy=dx/(x^2)$$
Integrating this, you get $$y^3/3= -1/x +c$$ where c is the constant of integration. Is there some other information given? By derivative of arctan, I guess you mean $$d(tan^-1(x))/dx =1/(1+x^2)$$

3. Dec 5, 2006

### singedang2

sorry, there was mistake in what was written in latex. it's fixed now.

4. Dec 5, 2006

### HallsofIvy

Staff Emeritus
I presume chaoseverlasting's response was to some incorrectly written equation which has been edited since the equation, as now given, is not separable.

I notice that the problem does not ask you to solve the differential equation, just to show that its solution must satisfy some property. But I wonder if there shouldn't be some initial value given? If the problem were to solve
$$f'(x)= \frac{1}{x^2+ f^2(x)}$$
with initial condition f(1)= 4, then since the right hand side is defined and differentiable at (1,4), the problem has a solution in some neighborhood of (1, 4) but clearly does not satisfy "$$|f(x)| \leq \frac{5 \pi}{4}< 4$$"

Last edited: Dec 5, 2006
5. Dec 5, 2006

### singedang2

oh right... f(1) = 1 i think this is the given value.

6. Dec 5, 2006

### singedang2

then does it mean now $$|f(x)| \leq \frac{5 \pi}{4}$$ ???
how can i show this using that differential equation?

7. Dec 5, 2006

any help????