First order nonlinear differential equation Help needed.

Click For Summary

Discussion Overview

The discussion revolves around solving a specific first-order nonlinear differential equation of the form 0 = a*[f(t)]^{z/(z-1)} + (-t+C)*f(t) + b*[df(t)/dt], where a, b, and C are constants and 0 < z < 1. Participants explore various methods and approaches to tackle this equation, including the use of integrating factors and substitutions.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Karen presents the nonlinear differential equation and requests assistance in solving it.
  • One participant suggests finding an integrating factor, providing a link to a step-by-step explanation.
  • Karen responds that the integrating factor technique is ineffective because it results in a function of both "t" and "f".
  • Another participant proposes simplifying the problem by working on a related, simpler differential equation and suggests a substitution method involving u=f^{1/2} to convert the equation.
  • Karen acknowledges the suggestion but notes that the negative value of z/(z-1) complicates the solution process.
  • Karen expresses difficulty in applying the proposed approach due to the problematic nature of the negative exponent.
  • A later reply introduces additional substitutions for different forms of the equation, suggesting u=f^{3/2} and u=f^{5/4} for further exploration.

Areas of Agreement / Disagreement

Participants do not reach a consensus on a solution method, and multiple competing approaches are presented. The discussion remains unresolved regarding the best way to tackle the original differential equation.

Contextual Notes

Participants express concerns about the implications of z/(z-1) being negative, which may affect the validity of certain solution methods. The discussion highlights the complexity of the problem and the challenges in applying standard techniques.

plushval
Messages
4
Reaction score
0
Hi All,

I have been trying to solve following nonlinear differential equation, but I couldn't solve it.

0 = a*[f(t)]^{z/(z-1)} + (-t+C)*f(t) + b*[df(t)/dt]

where a, b and C are constants and 0< z<1.

Could you please help me how to solve this nonlinear differential equation? I would really appreciate any suggestions. Many thanks, in advance!

Karen
 
Physics news on Phys.org
Thanks a lot for the suggestion, but unfortunately I could not make integrating factor technique work since integrating factors turns out to be a function of both "t" and "f". Any other suggestions?

Kind regards
 
I got a suggestion: When you get stuck, fall back and work on a simpler one first, then build it back up to the original problem:

f&#039;+f^{1/2}+(c-t)f=0

now, what happens if I let u=f^{1/2}

you can convert that to a DE in u right?. You know, f=u^2, f'=2uu'. You can do that? Ok, put all that back in, solve it for u, then of course f=u^2. Ok, how about:

f&#039;+f^{3/4}+(c-t)f=0

same dif. Let u=f^{1/4}. Now when you figure out what u is, then raise it to the fourth power to get f.

How about:

f&#039;+f^{n/m}+(c-t)f=0

How about when n/m is not rational?
 
Last edited:
I will definitely try to solve it in the way that you suggested jackmell. Thanks a lot!

As a quick note, z/(z-1) is negative; hence it may cause some problems.
 
As I said, z/(z-1) taking negative values is very problematic, therefore unfortunately, I could not solve the ODE using the approach you suggested, jackmall. I appreciate your time, though.

Any other suggestions would be highly appreciated.

Kind regards
 
If:

f&#039;+\frac{1}{f^{1/2}}+(c-t)f=0

then what happens if we let:

u=f^{3/2}

If:

f&#039;+\frac{1}{f^{1/4}}+(c-t)f=0

then what happens if we let:

u=f^{5/4}
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K