# Phase Line and long term behavior with initial value

• Feb 22nd 2014, 11:38 AM
goku900
Phase Line and long term behavior with initial value
Ok so i have the equation y'=y^2(y-3)(y-5)^3

I found the equilibrium positions to be y=0, y=3, y=5.

Does my phase diagram look right and would be correct in saying the solutions are nodes? Attachment 30216

The last part asks Describe the long term behavior of the solution to the above dierential equationwith initial condition y(0) = 4.

I'm not sure how to get started with this.
• Feb 22nd 2014, 12:06 PM
romsek
Re: Phase Line and long term behavior with initial value
deleted
• Feb 22nd 2014, 04:32 PM
goku900
Re: Phase Line and long term behavior with initial value
Deleted why ?
• Feb 22nd 2014, 07:28 PM
romsek
Re: Phase Line and long term behavior with initial value
because I wasn't confident it was correct. It's my reply that's deleted, not your question.
• Feb 22nd 2014, 07:44 PM
Prove It
Re: Phase Line and long term behavior with initial value
I'd probably start by solving the DE. It's separable...
• Feb 22nd 2014, 08:50 PM
romsek
Re: Phase Line and long term behavior with initial value
Quote:

Originally Posted by goku900
Ok so i have the equation y'=y^2(y-3)(y-5)^3

I found the equilibrium positions to be y=0, y=3, y=5.

Does my phase diagram look right and would be correct in saying the solutions are nodes? Attachment 30216

The last part asks Describe the long term behavior of the solution to the above dierential equationwith initial condition y(0) = 4.

I'm not sure how to get started with this.

Since you're not getting a ton of answers I'll write what my thinking was.

First off plot y' as a function of y

Attachment 30223

Now look at the stationary points 0, 3, and 5.

At y=0 a small excursion to the left leads to a positive velocity which leads back to 0.
A small excursion to the right also leads to positive velocity which further proceeds away from 0.

Thus 0 is an unstable equilibrium point.

At y=3 a small excursion to the left leads to positive velocity and back to 3.
A small excursion to the right leads to negative velocity and back to 3.

Thus 3 is a stable equilibrium point

At 5 you can see the equilibrium is unstable.

The phase diagram looks like

------------------->---0------->--------3----<-----5-------->

I'm not exactly sure what nodes means in the context of a single first order differential equation. I'd think all the equilibrium points have to be nodes.

You can solve this differential equation but Mathematica wasn't able to come up with a closed form solution for y in terms of x. You have to leave it implicit form.

I did solve it numerically and it exhibits the behavior described above, though while y=5 is unstable it's a pretty stable unstable, the velocity is very flat in this area and it takes a bit of a nudge sometimes to shake it off this point. Once y gets a tiny bit above 5 it takes off very rapidly towards infinity.

Looking at the phase diagram it's easy to see that at y[0]=4 the trajectory heads to and settles at y=3.