I am having trouble with a problem

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Homework Help Overview

The discussion revolves around finding equilibrium solutions for a second order nonlinear homogeneous differential equation, specifically u" + (u-1)u = 0. Participants are exploring various approaches to solve this equation and discussing related initial value problems involving Laplace transforms.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use a substitution method involving u' = v but encounters difficulties due to the lack of initial values. Another participant suggests a transformation that leads to a first order differential equation, noting that the resulting integral is complicated. Additionally, another post introduces a separate initial value problem involving Laplace transforms, highlighting challenges with partial fractions.

Discussion Status

Participants are actively engaging with the problem, offering different methods and transformations. There is no explicit consensus on a single approach, but some guidance has been provided regarding the use of substitutions and the nature of the integrals involved.

Contextual Notes

There are constraints related to the lack of initial values for the original differential equation and the complexity of the integrals involved. The discussion also notes the importance of boundary conditions in simplifying the problem.

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I am given the second order nonlinear homogeneous DE and I am supposed to find the equilibrium solutions for it.

u"+(u-1)u=0

I tried substuting u'=v but got stuck when I wasn't given initial values and couldn't solve for the constants after integrating.
 
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1weasel said:
I am given the second order nonlinear homogeneous DE and I am supposed to find the equilibrium solutions for it.

u"+(u-1)u=0

I tried substuting u'=v but got stuck when I wasn't given initial values and couldn't solve for the constants after integrating.

If you use the substitution:

[tex]u=y[/tex]
[tex]v=y'[/tex]
[tex]\frac{dv}{du}=\frac{y''}{y'}[/tex]

from which we have:

[tex]y=u[/tex]
[tex]y'=v[/tex]
[tex]y''=v\frac{dv}{du}[/tex]

You end up with a first order DE:

[tex]vdv=(1-u)udu[/tex]

Which is separable. The inverse substitution in the solution gives then a new first order equation in [itex]y'[/itex], which can be solved because it is also separable. However the integral is complicated. The end solution will be:

[tex]x=\int\frac{dy}{\sqrt{y^2-\frac{2}{3}y^3+K_1}}[/tex]

Perhaps you have boundary conditions which can make life easier, although I think it will stay an unpleasant integral. Hope this helps a bit.
 
Last edited:
There seems to be something wrong with the latex generation, so here my edit:

I rewrote the equation as:
[tex]\frac{d^2y}{dx^2}+y(y-1)=0[/tex]
 
I have tried to solve several times a Laplace transform with I.V.P's. There is what it looks like:

y" - y' -2y = F(t); where y(0) = 0, y'(0) = 0 and F(t) = 1 if 0<= t < 2
= t^2 + 1 if 2 <= t < 5
= t^2 + t if t => 5
I'm not having problem setting up but having trouble once I have everything on the right hand side. Unless I'm not using me brain the partial fractions look insane.
 
Please do not take over someone else's thread to ask a separate question. Start your own thread.

Solve the problem with F(t)= 1. Calculate the values of y and y' from that function at t= 2 and solve the initial value problem with F(t)= t2+ 1. Calculate the values of y and y' from that problem at t= 5 and solve the initial value problem with F(t)= t2+ t.
 

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