Solving Nonlinear ODE: y'y''=-1 - Get Help Now

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Homework Help Overview

The discussion revolves around solving a nonlinear ordinary differential equation (ODE) of the form y'y'' = -1. Participants are exploring various approaches to integrate and analyze the equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to manipulate the equation and suggests a potential integration approach. Some participants build on this by proposing integration of derived expressions and discussing the implications of different solutions.

Discussion Status

The discussion is active, with participants providing insights into the integration process and exploring multiple potential solutions. There is recognition of the existence of different forms of the solution, but no consensus has been reached on how to demonstrate the equivalence of these solutions.

Contextual Notes

Participants are navigating the complexities of nonlinear ODEs and considering the implications of different initial conditions or constants in their solutions. The nature of the problem suggests that assumptions about the behavior of y' may need further examination.

apolski
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Hello, i am trying to solve this nonlinear ODE

y'y''=-1

can someone help me?

p.s maybe 2y'y''=-2 => (y'y')'=-2...
 
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That's a good start. Now integrate ((y')^2)'=(-2).
 
οκ. [tex](y')^2=-2x+c[/tex] and

[tex]y(x)=\int\sqrt{-2x+c}\hspace{3}dx+c_2=[/tex][tex]-\frac{2\sqrt{2}}{3}\sqrt{(c_1 - x)^3}+c_2[/tex]

but i see that [tex]y(x)=\frac{2\sqrt{2}}{3}\sqrt{(c_1 - x)^3}+c_2[/tex] is sol'n too. How to show that?
 
Last edited:
If (y')^2=c-2x then y' is either +sqrt(c-2x) or -sqrt(c-2x). There are two solutions.
 
Thanx! o:)
 

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