Can This Nonlinear ODE with Exponential Terms Be Solved Using Power Series?

  • Context: Graduate 
  • Thread starter Thread starter MathematicalPhysicist
  • Start date Start date
  • Tags Tags
    Ode Thinking
Click For Summary

Discussion Overview

The discussion centers around the solvability of nonlinear ordinary differential equations (ODEs) involving exponential terms and other complex forms. Participants explore potential solutions using power series and inquire about the existence of special function forms for these equations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents the ODE y''(x) = y(x)e^{y'(x)} and attempts to find a solution via power series around x=0, expressing difficulty in deriving a general recurrence relation for the coefficients.
  • Another participant suggests that the ODE may be solvable in the form of the inverse function x(y) defined by an integral, though no further details are provided.
  • A participant introduces a different nonlinear ODE of the form y^{(n)}+(y^{(n-1)})^2+(y^{(n-2)})^3+\ldots + (y')^{n+1}+y^{n+2} = 0, questioning the feasibility of finding a power series solution and inquiring about the existence of a special function form solution.
  • Examples of specific cases for n=1, n=2, and n=3 are provided to illustrate the general case of the introduced ODE, highlighting the increasing complexity of the equations.

Areas of Agreement / Disagreement

Participants express differing views on the solvability of the presented ODEs, with no consensus reached on the existence of solutions or the methods to approach them. The discussion remains unresolved regarding the applicability of power series and special functions.

Contextual Notes

Participants note challenges in deriving recurrence relations and the potential convergence of the series, indicating that assumptions about the nature of the solutions may be incomplete or dependent on further definitions.

MathematicalPhysicist
Science Advisor
Gold Member
Messages
4,662
Reaction score
372
I thought today of the next DE:

[tex]y''(x) = y(x)e^{y'(x)}[/tex]

Not sure if it has applications, obviosuly I tried to find a solution via power series around x=0.

It seems tough to look for a general recurrence equation for the coefficients.
Here's what I have done so far.

[tex]y(x)=\sum_{n=0}^{\infty} a_n x^n[/tex]

[tex]y'(x)=\sum_{n=0}^{\infty} na_n x^{n-1}[/tex]

[tex]y''(x)=\sum_{n=0}^{\infty} n(n-1)a_n x^{n-2}[/tex]

equating:

[tex]\sum_{n=0}^{\infty} n(n-1)a_n x^{n-2} =\sum_{n=0}^{\infty} a_n x^n e^{\sum_{n=0}^{\infty} a_n n x^{n-1}}[/tex]

[tex]e^{a_1} e^{2a_2 x} e^{3a_3 x^2} \cdots = e^{a_1}[1+2a_1 x + \frac{(2a_1 x)^2}{2!}+\cdots]\cdot [1+3a_3 x^2 +\frac{(3a_3 x^2)^2}{2!}+\cdots]\cdot \cdots[/tex]

I am not sure if it even converges, is this equation known already, I am quite sure someone already thought of it.

Thanks in advance.
 
Last edited:
Physics news on Phys.org
Hi !
The ODE is solvable on the form of the inverse function x(y) as a special function defined by an integral :
 

Attachments

  • ODE Lambert.JPG
    ODE Lambert.JPG
    20.4 KB · Views: 454
Thanks.
 
Well, if I am already in the mood for non-ordinary DEs, I'll make this thread a thread with peculiar DEs I have in my mind.

Here's another one:

[tex]y^{(n)}+(y^{(n-1)})^2+(y^{(n-2)})^3+\ldots + (y')^{n+1}+y^{n+2} = 0[/tex]

Guessing a solution in the form of power series will be hard work (which I don't have time for right now), so does it have an specail function form solution?

P.S
n\geq 1
 
Maybe some examples if the general case isn't clear enough.

for n=1:

[tex]y'+y^2=0[/tex]

For n=2:

[tex]y''+(y')^2+y^3=0[/tex]

For n=3:

[tex]y'''+(y'')^2+(y')^3+y^4=0[/tex]

Etc.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
967
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 0 ·
Replies
0
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K