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Autonomous DE

  1. Jan 21, 2005 #1
    The problem is

    Consider the differential equation [tex] dy/dt=ay-b [/tex].
    a) Find the equilibrium solution [tex]y_e[/tex].
    b)Let [tex]Y(t)=y-y_e [/tex]; thus [tex]Y(t)[/tex] is the deviation from the equilbrium solution. Find the differential equation satisfied by [tex]Y(t)[/tex].

    For part a I am confused as to what is meant by [tex]y_e[/tex].
    The general solution is
    [tex]y=Ce^{at}+\frac{b}{a}[/tex]
    I thought that the equilibrium is just the value that will be approached as t increases without bound. So in this case it depends on the values of a. If a>0 then there is no equilibruim solution. How can I answer part a then?

    So without anywhere to go I made the assumption that [tex]y_e[/tex] is meant to mean the [tex]y(e)=y_e[/tex] in which case I come up with.
    [tex]Ce^{ae}+\frac{b}{a}=y_e[/tex]


    So if this in fact the equilibrium solution [tex]y_e[/tex] then for part b I have
    [tex]Y(t)=y-y_e[/tex]

    [tex]Y(t)=Ce^{at}+\frac{b}{a}-\left( Ce^{ae}+\frac{b}{a} \right)[/tex]

    [tex]Y(t)=ce^{at}-ce^{ae}[/tex]

    Which is really like any our first diff eq but in this case the [tex]\frac{b}{a}=-ce^{ae}[/tex] But the constant C makes our value for b/a unknown. Obviously I have something wrong here and I think its because I do not understand what the question is really asking.

    Thanks for any help
     
  2. jcsd
  3. Jan 21, 2005 #2

    dextercioby

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    This part

    is definitely wrong.You made a point.The general solution to the equation
    [tex] y(x)=Ce^{at}-\frac{b}{a} [/tex]

    must not diverge for asymptotic behavior...Which means that "a" must be negative...Find the equilibrium solution and then the function expressing the "deviation" from equilibrium...

    Daniel.

    PS.The diff.eq. will b very simple...
     
  4. Jan 21, 2005 #3
    Ok, well if know that a<0 then the equilibruim is going to be [tex]\frac{b}{a}[/tex]
    So then [tex]Y(t)=Ce^{at}[/tex]
    and
    [tex]\frac{dY}{dt}=ay[/tex]

    You said the diff.eq. will be b,I am not sure how you got that.

    sorry, I am trying to understand.

    Thanks
     
  5. Jan 21, 2005 #4

    dextercioby

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    That "b" was "be"... :tongue2: You got the picture. :smile: The diff.eq.may be put under the form which would definitely show the asymptotic behavior,namely make the substitution
    [tex] a\rightarrow -|a| [/tex]

    Daniel.
     
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