How can nonlinear ODEs be solved effectively?

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    Nonlinear Odes
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Discussion Overview

The discussion revolves around the effective methods for solving a specific nonlinear ordinary differential equation (ODE). Participants explore potential substitutions and transformations to facilitate the solution process, focusing on both the formulation of the equation and the implications of those transformations.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Homework-related, Mathematical reasoning

Main Points Raised

  • One participant presents a nonlinear ODE and requests assistance in solving it.
  • Another participant seeks clarification on the equation's formulation, pointing out a potential error in the notation regarding parentheses in the exponential function.
  • A suggestion is made to use the substitution \( u = \frac{x+y}{2x-y+1} \) to simplify the equation.
  • Following the substitution, a participant notes that the resulting equation is separable but mentions that the integral on the left side is not elementary.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the solution method, and multiple approaches are proposed, indicating that the discussion remains unresolved.

Contextual Notes

The discussion highlights potential ambiguities in the original equation's notation and the challenges associated with integrating the resulting expressions.

hatguy
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I need to solve the following ODE:

http://www.sosmath.com/CBB/latexrender/pictures/041ee1419e05bc0776451b294c1dcc0e.png

but i can't figure out a way to. Please help!
 
Last edited by a moderator:
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Regarding 3:
Did you mean
$$y'=e^{ \frac{x+y}{2x-y+1}}+\frac{3y-1}{3x+1}?$$
The absence of a closing parenthesis in the numerator of the argument of the exponential function makes your meaning unclear.
 
If it is indeed what Ackbach says, try letting $u = \dfrac{x+y}{2x-y+1}$.
 
Jester said:
If it is indeed what Ackbach says, try letting $u = \dfrac{x+y}{2x-y+1}$.

Nice! The result is separable. I get
$$\frac{e^{-u}}{u+1}\,u'=\frac{1}{3x+1}.$$

Of course, the integral on the left is not elementary. Oh, well.
 
Last edited:

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