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Math Help - Phase Line and long term behavior with initial value

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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? Phase Line and long term behavior with initial value-photo-4-.jpg


    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.
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    Re: Phase Line and long term behavior with initial value

    deleted
    Last edited by romsek; February 22nd 2014 at 12:08 PM.
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    Re: Phase Line and long term behavior with initial value

    Deleted why ?
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    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.
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    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 View Post
    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? Click image for larger version. 

Name:	photo (4).JPG 
Views:	3 
Size:	294.8 KB 
ID:	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

    Phase Line and long term behavior with initial value-clipboard01.jpg

    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.
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