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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.

Re: Phase Line and long term behavior with initial value

Re: Phase Line and long term behavior with initial value

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.

Re: Phase Line and long term behavior with initial value

I'd probably start by solving the DE. It's separable...

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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.