Find the Solution for x' = t + x/(1+x^2) with Initial Condition x(0) = 0

  • Thread starter Thread starter twoflower
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves solving the ordinary differential equation (ODE) x' = t + x/(1+x^2) with the initial condition x(0) = 0. This is situated within the context of non-linear differential equations, which may not have straightforward solutions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the bounds of the function x/(1+x^2) and how these might inform the solution. There are attempts to establish upper and lower bounds for x(t) based on these values. Questions arise regarding the feasibility of finding an analytical solution to the ODE.

Discussion Status

The discussion is ongoing, with some participants providing hints and exploring the implications of bounding the function. There is a recognition that non-linear ODEs often lack exact solutions, and some express dissatisfaction with this limitation while seeking alternative methods.

Contextual Notes

Participants note that the problem has not been covered in class, indicating a potential gap in knowledge or experience with this type of ODE. There is also mention of official results that differ from the bounds discussed, suggesting a need for clarification on the problem's expectations.

twoflower
Messages
363
Reaction score
0
Hi all,

I found this problem on web and though we haven't encountered this kind of problem in class I wonder how could it be done. Here it is:


Guess solution of the problem


[tex] x' = t + \frac{x}{1+x^2},\mbox{ x(0) = 0}[/tex]

ie. find functions [itex]\omega(t) \leq x(t) \leq \phi(t)[/itex] for each [itex]t[/itex] from domain of solution.

Well, I have no clue. I have the results here if anyone tries...

Thank you for hints!
 
Physics news on Phys.org
HINT: What are the largest and smallest possible values of the real function

[tex]\frac {x}{1+x^2}[/tex]
?
 
Tide said:
HINT: What are the largest and smallest possible values of the real function
[tex]\frac {x}{1+x^2}[/tex]
?

[tex] \frac{1}{2}[/tex]

and

[tex] -\frac{1}{2}[/tex] ?
 
Yes, and if you replace the "x" in the right hand side of the equation by those, can you then solve for x(t)?
 
HallsofIvy said:
Yes, and if you replace the "x" in the right hand side of the equation by those, can you then solve for x(t)?

Thank you HallsofIvy, so I wrote the bounds

[tex] x \leq \frac{t^2}{2} + \frac{t}{2}[/tex]

and

[tex] x \geq \frac{t^2}{2} - \frac{t}{2}[/tex]

It's ok, isn't it? Anyway, the official results say something slightly different...
 
twoflower said:
It's ok, isn't it? Anyway, the official results say something slightly different...


Ok, I have it. In the results, there is

[tex] \frac{t^2}{2} - t \leq x(t) \leq \frac{t^2}{2} + t[/tex]

which is just result of rougher bounds.
 
So can someone suggess a method for solving the original ODE analytically? I can't.
 
Non-linear differential equations, such as this one, tend NOT to have solutions that can be found exactly.
 
HallsofIvy said:
Non-linear differential equations, such as this one, tend NOT to have solutions that can be found exactly.

Hello Hall. How are you? May I say I'm not satisfied by this outcome? I know what you're thinkin': "why do I even bother; he's a pain in the . . .". Tell you what, suppose an asteriod was heading here and we had to find some analytical expression for this ODE in order to successfully deflect it. What progress could the combined intellect of the world muster to do so? I bet a whole dollar something could be done. :smile:
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 105 ·
4
Replies
105
Views
11K
Replies
3
Views
2K
Replies
5
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
Replies
4
Views
2K
Replies
4
Views
2K
Replies
6
Views
3K
Replies
7
Views
2K
Replies
8
Views
2K