Proving Bounds for f(x) using Differential Equations

[singedang2]

differential equation! urgent!

Homework Statement

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

Homework Equations

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!

 

[chaoseverlasting]

I don't 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)

 

[singedang2]

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

 

[HallsofIvy]

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"

 

[singedang2]

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

 

[singedang2]

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

 

[singedang2]

any help?