Stability Analysis of a Nonlinear Differential Equation

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Homework Help Overview

The discussion revolves around the stability analysis of a nonlinear differential equation represented by y' = Ay - By^2. Participants are tasked with identifying equilibrium solutions and classifying their stability.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore methods to find equilibrium solutions by setting y' to zero and factoring the equation. Questions arise regarding the correctness of identified solutions and the existence of additional solutions.

Discussion Status

Several participants have provided insights into finding equilibrium solutions, with some confirming the identification of solutions while others express uncertainty about the completeness of their findings. There is an ongoing exploration of the implications of these solutions on stability.

Contextual Notes

Participants are navigating through algebraic manipulations and questioning the assumptions related to the roots of the equation. There is acknowledgment of potential confusion stemming from earlier mathematical concepts.

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Homework Statement


Consider the differential equation given by

[tex]y' = Ay - By^2[/tex]

List the equilibrium solutions in increasing order and classify them as stable, semistable, or unstable.

Homework Equations


The Attempt at a Solution



Im just looking for a point in the right direction on this one. I guess my attempt would be to possibly factor out one of the y's on the right hand side. Then maybe move all the y's to the left side of the equation. Tell me... is this what I should be doing? The overall goal is to find up to 3 equilibrium solutions.
 
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Well you know at equilibrium that y' = 0 so...
 
Hootenanny said:
Well you know at equilibrium that y' = 0 so...

Thankyou.

[tex]0 = Ay - By^2[/tex]

[tex]By^2 = Ay[/tex]

[tex]By = A[/tex]

[tex]y = A/B[/tex]

the a/b is one equilibrium solution. Did I solve this the correct way or did I just luck out by doing this? Also how would I know that another existed (if it did)?
 
it should be [tex]y = \frac{a}{b}[/tex]. And look at the end behavior to look at stability(i.e. going away from eq. pt, converging to eq pt, or one part going away and one converging)
 
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I'm not sure, I'd have solve it like this;

[tex]\begin{array}{l}<br /> y(A - By) = 0\\<br /> \Rightarrow y = \left\{ 0,A/B\right\}<br /> \end{array}[/tex]

But you got one of the answers right :smile:
 
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Hootenanny said:
I'm not sure, I'd have solve it like this;

[tex]\begin{array}{l}<br /> y(A - By) = 0\\<br /> \Rightarrow y = \left\{ 0, A/B\right\}<br /> \end{array}[/tex]

But you got one of the answers right :smile:

omg its like basic algebra all over again. I am convinced that after so much math gets pushed inside your head, you start forgetting the earlier stuff. Thanks a ton.
 
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Ok, I'll be the one to throw it out there... where did the +/- come from? A-By=0 only if y is A/B, it's not if y=-A/B.
 
Wait, correction! The negative root isn't a solution. I really don't know what I was thinking! There are only two solutions (as the should be with a quadratic :mad:).

Edit: Shredder beat me to it. Sorry guys, this has happened twice now; I guess I should stop posting so late.
 
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