Solving ODE with Paper and Pencil: xy^2 - y = \frac {dy} {dx}

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Discussion Overview

The discussion revolves around solving the ordinary differential equation (ODE) given by \(\frac {dy} {dx} = x y^2 - y\) using paper and pencil methods. Participants explore various substitution techniques and the classification of the ODE.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant shares a solution obtained using Mathematica's DSolve function, indicating the solution is \(y(x) = \frac {1} {1 + x + C e^{x}}\), but expresses uncertainty about solving it manually.
  • Another participant suggests a substitution \(y(x) = \frac{1}{u(x)}\) and believes this will lead to a more manageable ODE.
  • A third participant acknowledges unfamiliarity with the substitution method but finds it elegant after researching it.
  • A fourth participant contextualizes the problem by noting its form as a Bernoulli ODE or a Ricatti equation without a constant term, proposing the substitution \(u(x) = y^{1-p}\) to reduce it to a first-order linear ODE solvable by integrating factors.

Areas of Agreement / Disagreement

Participants do not reach a consensus on a single method for solving the ODE manually, as multiple approaches and substitutions are proposed, indicating a variety of perspectives on the problem.

Contextual Notes

The discussion includes various assumptions about the applicability of substitution methods and the classification of the ODE, but these assumptions remain unresolved.

Neoma
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[tex]\frac {dy} {dx} = x y^2 - y[/tex]

I used Mathematica's DSolve function and found the correct answer:
[tex]y(x) = \frac {1} {1 + x + C e^{x}}[/tex]

However, I don't have any idea what method to use to solve it with pencil and paper...
 
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I cheated,i know;i probably wouldn't have seen it,if you hadn't provided the answer.

Make the substitution:

[tex]y(x)=\frac{1}{u(x)}[/tex]

I believe you'll like the ODE that comes out.

Daniel.
 
I wasn't familiar with the substitution method yet, so I looked it up after reading your post and it looks quite elegant :)

Thanks!
 
Just to put this problem in a general context, it's form is:

[tex]\frac {dy} {dx} = a(x) y + b(x) y^p[/tex]

Which is a Bernoulli ODE (or a Ricatti with no constant term).

The substitution:

[tex]u(x) = y^{1-p}[/tex]

reduces this to a first order linear ODE which can be solved in the usual way via an integrating factor.
 

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