Consider the differential equation $x'=x^2-9$

a. find the stability type of each fixed point

To find the fixed points, I set this equal to $0$, right? Would someone mind explaining why I do this? I don't really understand the concept.

So I get the fixed point as being $x=3$ and $x=-3$.

Checking the stability:

For $x=3$:
$x=2$ results in $-5$ (left)
$x=4$ results in $7$ (right)


For $x=-3$:
$x=-4$ is positive (left)
$x=-2$ is negative (right)


How do I determine their stability exactly?

b. sketch the phase portrait on the line

c. Sketch the graph of $x(t)=ϕ(t;x_0)$ in the $(t,x)$-plane for several representative initial conditions $x_0$.

  • 1
    $\begingroup$ You mean $x' = x^2-9$? $\endgroup$
    – graydad
    Nov 7, 2014 at 16:48
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    $\begingroup$ Fixed points are points where the solution to the differential equation is, well, fixed. That is, it doesn't move (i.e. doesn't change with respect to $t$). If it isn't moving, its derivative is zero. If its derivative $x'$ is zero, this means that $x^2 - 9$ is zero as well. $\endgroup$
    – BaronVT
    Nov 7, 2014 at 16:51
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    $\begingroup$ You might also say a little more about your thoughts on the other parts of this question; what have you tried, or what similar problems have you completed, and why is this one different? (or maybe you have no idea what a phase portrait is, but you should say so) $\endgroup$
    – BaronVT
    Nov 7, 2014 at 16:54
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    $\begingroup$ @MathMajor yes both are identical. I used the "prime" notation in my class but I know "dot" notation is fairly standard. $\endgroup$
    – graydad
    Nov 7, 2014 at 16:56
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    $\begingroup$ @MathMajor Also, remember $x$ is a function with input $t$, so you aren't checking $x = -2$, you are checking $t = -2 \implies$ checking $x'(-2)$. Do you know how to sketch a phase portrait? $\endgroup$
    – graydad
    Nov 7, 2014 at 17:01

2 Answers 2


To find the fixed points, we set $x' = 0$ and solve, yielding:

$$x' = x^2 -9 = 0 \implies x_{1,2} = \pm~3$$

To test stability, you can follow Paul's Online Notes, by picking values around the critical points and noting the sign of the derivative $x'$.

If we draw a phase line, we get (note that $+3$ is unstable and $-3$ is stable):

enter image description here

If we draw a direction field plot and then superimpose solution curves on it, we have (compare the two critical points to the phase line and look at each (purple) solution curve):

enter image description here

Note, for the solution curves (the direction field plot shows many examples), you can take several examples for different ICs, for example (just as the plot shows).

$$x' = x^2 - 9, x(0) = 1 \implies x(t) = \dfrac{6-3 e^{6 t}}{e^{6 t}+2}$$

What happens to $x(t)$ as $t$ approaches infinity? It approaches $-3$.


Remember, $x'(t)$ tells us the rate of change of $x(t)$. You want to set it equal to zero because that indicates a point where $x(t)$ is not changing (that is to say, it is fixed). To determine stability, check values of $x'(t)$ on the left and right of your fixed points. For example, for the fixed point $t = -3$, check the value of $x'(-4)$ and $x'(-2)$. If $x'(t)$ is positive to the left of $x'(-3)$ and negative to the right of $x'(-3)$ then $t=-3$ is stable. If the left and right are the same sign, you have semi-stability. If the left side is negative while the right is positive, you have instability.

Hopefully this picture illustrates that solutions move towards $x(-3)$ and away from $x(3)$


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