# Thread:

1. ## Problem With Stability Theorem

Okay, I have the following autonomous differential equation:

$dx/dt = f(x) = x^2-5x+6$

I found the steady states of the equation of be x*1=2 and x*2=3

When I sub them into $f'(x^*) = 2x-5$, I get $f'(x^*1) = -1$ and $f'(x^*2)= 1$.

Now I was told that the stability theorem states that if $f'(x^*) < 0$, then x* is stable and when $f'(x^*) > 0$, x* is unstable, however, when I create my phase line diagram is shows that x*2 is stable.

Could someone shed some light on this?

Thanks,

JS

2. Originally Posted by jschlarb
Okay, I have the following autonomous differential equation:

$dx/dt = f(x) = x^2-5x+6$

I found the steady states of the equation of be x*1=2 and x*2=3

When I sub them into $f'(x^*) = 2x-5$, I get $f'(x^*1) = -1$ and $f'(x^*2)= 1$.

Now I was told that the stability theorem states that if $f'(x^*) < 0$, then x* is stable and when $f'(x^*) > 0$, x* is unstable, however, when I create my phase line diagram is shows that x*2 is stable.

Could someone shed some light on this?

Thanks,

JS
Here's the direction field- it shows that $x = 2$ is stable and $x = 3$ unstable.
Attached Thumbnails

3. I think I found my problem: When I made a general shape of the graph, I made it so it curved down instead of curve up :S.

Thanks!

4. I also have another part of the question I need help with:

It asks to use separation of variable to determine the equation explicitly, however when I integrate it, I get an equation that doesn't make sense:

$1=Ke^t$

I separated the variables to get an equation with which I needed to use integration by parts. I determined that A=1 and B=-1 to which i got the final equation to be

$ln|x-2|-ln|x-3|+c = t+c$

Taking the e function to everything and simplifying got me that. Could there have been a mistake in my integration (I apologize, I'm not too familiar with the math functions on this site so I wasn't able to show you what I did).

5. Originally Posted by jschlarb
I also have another part of the question I need help with:

It asks to use separation of variable to determine the equation explicitly, however when I integrate it, I get an equation that doesn't make sense:

$1=Ke^t$

I separated the variables to get an equation with which I needed to use integration by parts. I determined that A=1 and B=-1 to which i got the final equation to be

$ln|x-2|-ln|x-3|+c = t+c$

Taking the e function to everything and simplifying got me that. Could there have been a mistake in my integration (I apologize, I'm not too familiar with the math functions on this site so I wasn't able to show you what I did).
No - you're good to go. Put the constants together and call them $\ln c$ (it cleans things up a bit.) Then solve for x

$\ln \left| \frac{x-2}{x-3}\right| = \ln c + t$

and solving for x gives

$x = \frac{3 c e^t - 2}{ c e^t - 1}$

You can see if $c = 0$ you have your first critical point whereas if you let $c = \frac{1}{\bar{c}}$ then
$x = \frac{3e^t - 2\bar{c}}{ e^t - \bar{c}}$

and letting $\bar{c} = 0$ gives your second.
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