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Math Help - The long term behaviour of the solution to the differential equation

  1. #1
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    Question The long term behaviour of the solution to the differential equation

    I attempted this differential equation but couldn't find the answer to part (c) of the question!
    This question is about the differential equation dy/dt = y^2 - 4y + 2.
    (a) Find all equilibrium points. Determine whether each equilibrium is a sink, a source, or a node.
    (b) Sketch the phase line
    (c) Describe the long term behaviour of the solution to the differential equation that satisfies the initial condition:
    (i) y(0) = 0;
    (ii) y(0) = 1;

    For part (a) I found the equilibrium solutions as y=2+sqrt(2) and y = 2-sqrt(2) with y=2+sqrt(2) being a source and y=2-sqrt(2) being a sink. For part (b) I made the phase line having both the equilibrium solutions and the directional arrows. I however couldn't find the answer to part (c) of the question. I couldn't find anything similiar to part (c) either. Help please. Thanks!
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    Quote Originally Posted by Skelly View Post
    I attempted this differential equation but couldn't find the answer to part (c) of the question!
    This question is about the differential equation dy/dt = y^2 - 4y + 2.
    (a) Find all equilibrium points. Determine whether each equilibrium is a sink, a source, or a node.
    (b) Sketch the phase line
    (c) Describe the long term behaviour of the solution to the differential equation that satisfies the initial condition:
    (i) y(0) = 0;
    (ii) y(0) = 1;

    For part (a) I found the equilibrium solutions as y=2+sqrt(2) and y = 2-sqrt(2) with y=2+sqrt(2) being a source and y=2-sqrt(2) being a sink. For part (b) I made the phase line having both the equilibrium solutions and the directional arrows. I however couldn't find the answer to part (c) of the question. I couldn't find anything similiar to part (c) either. Help please. Thanks!
    You could always solve the DE and consider the solution as t--> oo ......

    Alternativley, if the long term behaviour is that y --> constant, then dy/dt --> 0 => y^2 - 4y + 2 --> 0 .....

    Since y(0) = 0 is under y = 2 - sqrt{2}, it looks to me that y --> 2 - sqrt{2} for this initial condition.

    Since y(0) = 1 is above y = 2 - sqrt{2} and below y = 2 + sqrt{2}, it looks to me like y --> 2 + sqrt{2} for this initial condition.
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    Quote Originally Posted by mr fantastic View Post
    You could always solve the DE and consider the solution as t--> oo ......

    Alternativley, if the long term behaviour is that y --> constant, then dy/dt --> 0 => y^2 - 4y + 2 --> 0 .....

    Since y(0) = 0 is under y = 2 - sqrt{2}, it looks to me that y --> 2 - sqrt{2} for this initial condition.

    Since y(0) = 1 is above y = 2 - sqrt{2} and below y = 2 + sqrt{2}, it looks to me like y --> 2 + sqrt{2} for this initial condition.
    *Ahem* Having used my synapse, I will edit my answers and change one of them:

    Since y(0) = 0 is under y = 2 - sqrt{2}, and dy/dt > 0 when t = 0 and y = 0, it looks to me that y --> 2 - sqrt{2} for this initial condition.


    Since y(0) = 1 is above y = 2 - sqrt{2} and below y = 2 + sqrt{2}, and dy/dt < 0 when t = 0 and y = 1, it looks to me like y --> 2 - sqrt{2} for this initial condition
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