- #1
Townsend
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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
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