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OK, this is a weak question, but I'm a bit rusty on differential equations, so here it is:

I need to find the general solution of x'=ax+3 and then find the equilibrium points as well as which points are equilibria sinks and sources (a is a parameter).

I found the general solution of x(t). It was: $x(t)=Ce^{at}-\frac{3}{a}$

Now, from the one or two lines in the text about equilibrium points, I'm seeing that that will be a "value" that makes x(t) = 0. Though, the way it's worded in the text makes it seem like a choice of "a" or "C" will be what makes an equilibrium solution.

This is where I'm confused. Do I solve the general solution for t by setting x = 0? Do I solve the original equation finding setting x' to 0 and finding x (this is what I saw on some website).

I'm just not sure what I'm supposed to do to find the equilibrium points, and then how to decide which are sinks and which are sources.

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1 Answer 1

up vote 5 down vote accepted

Your differential equation is $x' = ax+3$. An equilibrium solution is one for which $x'=0$ along the solution. As a counter-example, $x(t)=0$ (the zero function) is not an equilibrium solution because although $x'=0$ it is not true that $a(0)+3=x'$. The zero function is not a solution to the given differential equation. We need two things:

1.) the proposed solution has the property $x'=0$

2.) the proposed solution is in fact a solution (when you plug it into the DEQn it works)

Therefore, $x'=ax+3=0$ yields $x = -3/a$ as the equilbrium solution.

For more complicated differential equations the equilibrium solutions can be more interesting. Here this solution is actually the exceptional case in your general solution if you derived it by separation of variables. Notice,

$$ \frac{dx}{dt} = ax+3 $$

yields

$$ \int \frac{dx}{ax+3}= \int dt $$

provided $ax+3 \neq 0$. Integrating,

$$ \frac{1}{a}\ln |ax+3| = t+c $$

algebra,

$$ |ax+3| = e^{at+ac} =e^{ac}e^{at} $$

Hence,

$$ ax+3 = \pm e^{ac}e^{at} $$

Or, $x = Ce^{at} - 3/a$ where $C = \pm e^{ac}/a$. Notice that $C$ defined in this way ought not be zero. It is interesting that $C=0$ brings us the equilibrium solution.

Often we find equilibrium solutions between other classes of solutions. For example, see the logistic equation $P' = P(1-P/C)$. The equilibrium solution is attained by the constant solution $P=C$ whereas there are two other solution classes which asymptotically approach the equilibrium either from below or from above.

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Thanks! I actually finished this up after finding a nice website with a clear description of the equilibrium solution (my text is awful), but it's nice to confirm that what I did worked! So thanks a bunch! –  user25326 Aug 30 '12 at 5:45
    
glad to almost help :) –  James S. Cook Aug 30 '12 at 5:49

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