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Math Help - First order DE

  1. #1
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    First order DE

    Let

    \frac{dx}{dt}=-x-x^{2}+y^{2}<br />
and<br />
\frac{dy}{dt}=-y-2xy.

    I have shown that the two equations are the real and imaginary parts of a single differential equation for the complex variable
    z(t), \frac{dz}{dt}=-z-z^{2}.

    I have evaluated the differential equation for z(t) to get

    t+C=\ln\left(\frac{z+1}{z}\right).

    However, I need to determine the solution to the DE with initial condition z=i at t=0. I then need to use this to show that the solution of the original equations for x(t) and y(t) with initial conditions
    x(0)=0 and y(0)=1 has both x(t) and y(t) approaching \exp\left(\frac{-t}{2}\right)/2 as t\rightarrow\infty.

    Help!
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  2. #2
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    Quote Originally Posted by Cairo View Post
    Let

    \frac{dx}{dt}=-x-x^{2}+y^{2}<br />
and<br />
\frac{dy}{dt}=-y-2xy.

    I have shown that the two equations are the real and imaginary parts of a single differential equation for the complex variable
    z(t), \frac{dz}{dt}=-z-z^{2}.

    I have evaluated the differential equation for z(t) to get

    t+C=\ln\left(\frac{z+1}{z}\right).

    However, I need to determine the solution to the DE with initial condition z=i at t=0. I then need to use this to show that the solution of the original equations for x(t) and y(t) with initial conditions
    x(0)=0 and y(0)=1 has both x(t) and y(t) approaching \exp\left(\frac{-t}{2}\right)/2 as t\rightarrow\infty.

    Help!

    From your solution

     <br />
\frac{z+1}{z} = k e^t<br />
    and with your IC

     <br />
\frac{z+1}{z} = \frac{i+1}{i} e^t = (1 - i) e^t<br />

    solving for z gives

     <br />
z = \frac{1}{\left( e^t - 1 \right) - i e^t} = \frac{\left( e^t - 1 \right) + i e^t}{\left( e^t - 1 \right) ^2 + e^{2t}}<br />

    giving

     <br />
x = \frac{\left( e^t - 1 \right) }{\left( e^t - 1 \right) ^2 + e^{2t}}<br />
,  <br />
y = \frac{ e^t}{\left( e^t - 1 \right) ^2 + e^{2t}}<br />

    the solutions of your DE noting that  x(0) = 0, y(0) = 1. Now when t \to \infty then x and y approach

     <br />
x,y \to \frac{e^t}{2 e^{2t}} = \frac{e^{-t}}{2}<br />
(not quite what you got but close.)
    Last edited by Jester; March 5th 2010 at 06:31 AM. Reason: fixed latex
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  3. #3
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    Quote Originally Posted by Danny View Post
    From your solution

     <br />
\frac{z+1}{z} = k e^t<br />
    and with your IC

     <br />
\frac{z+1}{z} = \frac{i+1}{i} e^t = (1 - i) e^t<br />

    solving for z gives

     <br />
z = \frac{1}{\left( e^t - 1 \right) - i e^t} = \frac{\left( e^t - 1 \right) + i e^t}{\left( e^t - 1 \right) ^2 + e^{2t}}<br />

    giving

     <br />
x = \frac{\left( e^t - 1 \right) }{\left( e^t - 1 \right) ^2 + e^{2t}}<br />
,  <br />
y = \frac{ e^t}{\left( e^t - 1 \right) ^2 + e^{2t}}<br />

    the solutions of your DE noting that  x(0) = 0, y(0) = 1. Now when t \to \infty then x and y approach

     <br />
x,y \to \frac{e^t}{2 e^{2t}} = \frac{e^{-t}}{2}<br />
(not quite what you got but close.)
    Thanks for this.

    How did you get

     <br />
x,y \to \frac{e^t}{2 e^{2t}} = \frac{e^{-t}}{2}<br />

    from

     <br />
x = \frac{\left( e^t - 1 \right) }{\left( e^t - 1 \right) ^2 + e^{2t}}<br />
,  <br />
y = \frac{ e^t}{\left( e^t - 1 \right) ^2 + e^{2t}}<br />
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  4. #4
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    Quote Originally Posted by Cairo View Post
    Thanks for this.

    How did you get

     <br />
x,y \to \frac{e^t}{2 e^{2t}} = \frac{e^{-t}}{2}<br />

    from

     <br />
x = \frac{\left( e^t - 1 \right) }{\left( e^t - 1 \right) ^2 + e^{2t}}<br />
,  <br />
y = \frac{ e^t}{\left( e^t - 1 \right) ^2 + e^{2t}}<br />
    I'll do x, y follows similarly. Expanding x gives

     <br />
x = \frac{e^t - 1}{ 2 e^{2t} - 2 e^t + 1}<br />

    Hand waving. In the numerator, as t \to \infty, the e^t is more dominate than the -1. In the denominator, the 2e^{2t} is more dominate than 2e^t and 1. Hence

    x \to \frac{e^t}{2e^{2t}} as t \to \infty.

    You could be a little more formal using limits but this gets the idea across.
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  5. #5
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    Thanks. It suddenly dawned on me to use L'Hopital's rule and the result follows.

    Thanks for your help.
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