Solve Non-Linear ODE with a Substitution | Step-by-Step Guide

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    Non-linear Ode
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Discussion Overview

The discussion revolves around solving a non-linear ordinary differential equation (ODE) of the form xy'' + y' + (y')^3 = 0. Participants explore potential substitutions to simplify the equation and discuss the feasibility of finding an exact analytical solution.

Discussion Character

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

Main Points Raised

  • One participant requests assistance with a non-linear ODE and seeks a suitable substitution.
  • Another participant questions the existence of a "good substitution," noting that many non-linear differential equations do not have exact solutions.
  • A different participant argues that since the problem is from a text, it implies that an exact analytical solution is expected to exist.
  • One participant suggests a substitution of z = y' to reduce the order of the equation, transforming it into a first-order separable equation.

Areas of Agreement / Disagreement

There is no consensus on the existence of a "good substitution" or the solvability of the ODE. Some participants express skepticism about finding an exact solution, while others believe it is possible due to the context of the problem.

Contextual Notes

Participants note the presence of y' twice in the equation without y, which may affect the approach to solving it. The discussion includes assumptions about the solvability of non-linear ODEs and the implications of the problem's origin in a text.

SeReNiTy
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Hi guys, just wondering if you can give a hand on a non-linear ODE, all you guys need to do if give me a good substituion to try...

xy''+y'+(y')^3 = 0

where ' = d/dx
 
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Do you have any reason to believe there is a "good substitution"?

Almost all non-linear differential equations cannot be solved exactly.
 
Because this is a problem in a text, so i believe when thye say find the exact analytical solution, they must mean one exists...
 
Okay, that's a good reason! Also I just noticed that you have y' twice and no y. If you let z= y', you can reduce the order of the equation:
xy''+y'+(y')^3 = 0 becomes xz'+ z+ z3= 0, a first order, separable equation. Solve for z and let y'= z.
 

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