Autonomous Differential Equations: Equilibrium Solution and Deviation Equation

[Townsend]

The problem is

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

For part a I am confused as to what is meant by y_e.
The general solution is
y=Ce^{at}+\frac{b}{a}
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 y_e is meant to mean the y(e)=y_e in which case I come up with.
Ce^{ae}+\frac{b}{a}=y_e


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

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

Y(t)=ce^{at}-ce^{ae}

Which is really like any our first diff eq but in this case the \frac{b}{a}=-ce^{ae} 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

 

[dextercioby]

This part

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


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

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

Y(t)=ce^{at}-ce^{ae}

Which is really like any our first diff eq but in this case the \frac{b}{a}=-ce^{ae} 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.

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

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...

 

[Townsend]

Ok, well if know that a<0 then the equilibruim is going to be \frac{b}{a}
So then Y(t)=Ce^{at}
and
\frac{dY}{dt}=ay

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

sorry, I am trying to understand.

Thanks

 

[dextercioby]

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

Daniel.