Q How to Solve Nonlinear ODEs with Two Variables and Homogeneous Equations?

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    Nonlinear Odes
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Discussion Overview

The discussion revolves around solving two nonlinear ordinary differential equations (ODEs) involving two variables. Participants explore various substitutions and transformations to simplify the equations, focusing on theoretical approaches and mathematical reasoning.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests substituting $u=\ln(y)$ for the first ODE, leading to a quadratic form in $u'$ and proposes finding $u'$ as $\frac{-x \pm \sqrt{x^{2}+4u}}{2}$.
  • Another participant mentions that the substitution $u=y^{2}$ for the second ODE transforms it into a Ricatti equation.
  • A further update indicates that an additional substitution $v=-1+\sqrt{1+4u}$ may render the first equation separable.
  • One participant proposes differentiating the expression provided by another participant, leading to a new equation $u''(2u'+x) = 0$, suggesting two cases to consider.
  • Reiteration of the substitution $u = y^2$ and introducing $t = x^4$ is mentioned, with the claim that the new equation should be homogeneous.

Areas of Agreement / Disagreement

Participants present multiple approaches and transformations for the ODEs, but there is no consensus on a single method or solution. The discussion remains unresolved with various competing views and methods proposed.

Contextual Notes

Some participants express uncertainty about the next steps after certain substitutions, and there are indications of missing assumptions or dependencies on specific definitions related to the transformations used.

hatguy
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I need to solve 2 ODEs:

1. http://www.sosmath.com/CBB/latexrender/pictures/7b213e6c9e4d5fd9d92877694610ac22.png
2. http://www.sosmath.com/CBB/latexrender/pictures/528f96046147932945da54b7a47f97a9.pngbut i can't figure out a way to. Please help!
 
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Regarding 1:

If you let $u=\ln(y)$, then the equation
$$\ln(y)=\left( \frac{y'}{y}\right)^{\!\!2}+x\,\frac{y'}{y}$$
reduces to
$$u=(u')^{2}+xu'.$$
If you view this is a quadratic in $u'$, you can find that
$$u'=\frac{-x \pm \sqrt{x^{2}+4u}}{2}.$$
Not sure where you could go from here. You could try to make it exact.

Regarding 2:

The substitution $u=y^{2}$ renders the equation Ricatti. Have fun with that!
 
Further update on Number 1: the additional substitution $v=-1+\sqrt{1+4u}$ renders the equation separable, I think.
 
Q1 Differentiate what Ackbach has giving $u''(2u'+x) = 0$ - two cases to consider.

Q2 As Ackbach said let $u = y^2$, further let $t = x^4$. Your new equation should be homogeneous.
 
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Jester said:
Q1 Differentiate what Ackbach has giving $u''(2u'+x) = 0$ - two cases to consider.

Q2 As Ackbach said let $u = y^2$, further let $t = x^4$. Your new equation should be homogeneous.

Thanks, everyone, I found my solutions. And yes, a left bracket shouldn't be where it is now.
 

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