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Math Help - Simultaneous DE

  1. #1
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    Simultaneous DE

    dx/dt=-ax

    dy/dt=ax-by

    where a and b are constants. Solve this system subject to x(0)=x_0 (x "not") and y(0)=y_0 (y "not")

    Is it possible to substitute -dx/dt into equation 2 or do I have to solve equation one first to find the solution of x first?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by daskywalker View Post
    dx/dt=-ax

    dy/dt=ax-by

    where a and b are constants. Solve this system subject to x(0)=x_0 (x "not") and y(0)=y_0 (y "not")

    Is it possible to substitute -dx/dt into equation 2 or do I have to solve equation one first to find the solution of x first?
    solve the first DE for x (it is a separable DE), and plug that into the second DE.

    (and it is x- and y- naught, as in zero)
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  3. #3
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    ok so for the first DE I got x=Ke^-at where K is the constant e^c (from integration) How do I substitute the condition x(0)=x_0? Or do I substitute the whole expression into equation 2?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by daskywalker View Post
    ok so for the first DE I got x=Ke^-at where K is the constant e^c (from integration) How do I substitute the condition x(0)=x_0? Or do I substitute the whole expression into equation 2?
    x(0) = x_0 means, when t = 0,~x = x_0

    We have x = Ke^{-at}. applying the initial condition, we see that x_0 = Ke^0 = K

    Thus, x = x_0e^{-at}. Plug this expression in the second equation
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