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Math Help - Coupled differential equations

  1. #1
    Member Maccaman's Avatar
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    Coupled differential equations

    I really need help solving these coupled differential equations

     \frac{dZ}{dt} = -aZ

     \frac{dY}{dt} = cZ - gY

    where a = b+c+g and a,b,c,g are all constants.

    The initial conditions are

     Z(0) = Z_0 , and  Y(0) = Y_0 ,
    I know that  Z(t) = e^{-at} , but im having trouble with finding Y(t)

    This is a past exam question so I have the answer already, but I need to know exactly how to do it for a homework question I have.
    Here is the solution

     Y(t) = Y_0 e^{-gt} + \frac{c*Z_0*(e^{-gt} - e^{-at}) }{a - g}
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  2. #2
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    I only remember a few steps in these kinds of problems...
    BTW Z = Z_0 e^{-at} not just e^{-at}
    Y = PI + CF = particular integral + complementary function I believe
    You're given
    Dy = cZ - gY where D = differentiation operator so
    (D+g) Y = cZ
    PI (or CF) is given by (D+g) Y = 0 i.e. PI = const * e^{-gt}
    CF (or PI) is given by (D+g) Y = cZ = he^{-at} where h = cZ_0
    Assume CF= ve^{-wt} then (D+g) Y = v(g-w)e^{-wt} , v w const
    Eqauating powers of t, w = a so CF = ve^{-at}
    v(g-a)e^{-at} = cZ_0e^{-at} so v = cZ_0/(g-a)
    Now Y = PI + CF = ke^{-gt} + cZ_0/(g-a) e^{-at}
    Almost there, hopefully someone more knowledgeable will correct my missteps here
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Maccaman View Post
    I really need help solving these coupled differential equations

     \frac{dZ}{dt} = -aZ

     \frac{dY}{dt} = cZ - gY

    where a = b+c+g and a,b,c,g are all constants.

    The initial conditions are

     Z(0) = Z_0 , and  Y(0) = Y_0 ,
    I know that  Z(t) = e^{-at} , but im having trouble with finding Y(t)

    This is a past exam question so I have the answer already, but I need to know exactly how to do it for a homework question I have.
    Here is the solution

     Y(t) = Y_0 e^{-gt} + \frac{c*Z_0*(e^{-gt} - e^{-at}) }{a - g}
    You're very close for the solution to Z(t). Z(t)={\color{red}C}e^{-at}

    Since Z(0)=Z_0,~Z_0=C\implies C=Z_0

    So Z(t)=Z_0e^{-at}

    Now plug your solution for Z into the differential equation you have for Y:

    \frac{\,dY}{\,dt}=cZ_0e^{-at}-gY

    Rewrite the equation to see that its a linear equation:

    \frac{\,dY}{\,dt}+gY=cZ_0e^{-at}

    The integrating factor is \varrho(t)=e^{\int g\,dt}=e^{gt}

    Thus, we see that \frac{d}{\,dt}\left[Ye^{gt}\right]=cZ_0e^{(g-a)t}

    Integrating both sides yields:

    Ye^{gt}=\frac{cZ_0}{g-a}e^{(g-a)t}+C\implies Y=\frac{cZ_0}{g-a}e^{-at}+Ce^{-gt}

    Applying the initial condition Y(0)=Y_0, we see that Y_0=\frac{cZ_0}{g-a}+C\implies C=Y_0-\frac{cZ_0}{g-a}

    Thus, Y=\frac{cZ_0}{g-a}e^{-at}+\left[Y_0-\frac{cZ_0}{g-a}\right]e^{-gt}

    Therefore,

    \color{red}\boxed{Y(t)=Y_0e^{-gt}+\frac{cZ_0\left[e^{-gt}-e^{-at}\right]}{a-g}}

    Does this make sense?

    --Chris
    Last edited by Chris L T521; September 18th 2008 at 01:56 PM. Reason: oops...typos. never copy and paste! >.<
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  4. #4
    Member Maccaman's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    You're very close for the solution to Z(t). Z(t)={\color{red}C}e^{-at}

    Since Z(0)=Z_0,~Z_0=C\implies C=Z_0

    So Z(t)=Z_0e^{-at}
    What I originally wrote was a typo. I did mean to type Z(t)=Z_0e^{-at}

    But thanks for providing me with the details for obtaining Y(t). It makes perfect sense. You
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  5. #5
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    another way

    or the system:

    <br />
\frac{dx}{dt}=5x+3y

    \frac{dy}{dt}=x+7y<br /> <br />

    You determined the eigenvalues and eigenvectors. Thus the solution is:

    <br />
\left(<br />
\begin{array}{c} x(t) \\y(t)\end{array}\right)<br />
=k_1e^{4t}<br />
     \left(\begin{array}{c} 3    \\ -1 \end{array}\right)+k_2e^{8t}<br />
     \left(\begin{array}{c} 1    \\ 1  \end{array}<br />
\right)<br />

    Right?

    Substituting the initial conditions:

    <br /> <br />
\left(\begin{array}{c} 5 \\ 1 \end{array}\right)=<br />
k_1\left(\begin{array}{c}3 \\ -1 \end{array}\right)+<br />
k_2\left(\begin{array}{c}1 \\ 1  \end{array}\right)<br />

    Now just figure out what the constants are. You know how to read this matrix stuf huh? That last one is just:

    <br />
 5=3k_1+k2
     1=-k_1+k_2<br />

    Same dif for reading the other one.

    Edit: Oh yea:

    Welcome to PF and

    "Differential Equations" by Blanchard, Devaney, and Hall is one DE book I use.
    Last edited by CaptainBlack; November 15th 2009 at 07:22 AM. Reason: I hope to correct defective LaTeX
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  6. #6
    MHF Contributor
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    Quote Originally Posted by ajbriggs View Post
    or the system:

    <br /> <br />
\frac{dx}{dt}=5x+3y<br /> <br />
\frac{dy}{dt}=x+7y<br /> <br />
    You determined the eigenvalues and eigenvectors. Thus the solution is:

    <br />
\left(\begin{array}{c} x(t) \\y(t)\end{array}\right)=k_1e^{4t}<br />
\left(\begin{array}{c} 3 \\ -1 \end{array}\right)+k_2e^{8t}<br />
\left(\begin{array}{c} 1 \\ 1 \end{array}\right)<br />

    Right?

    Substituting the initial conditions:

    <br /> <br />
\left(\begin{array}{c} 5 \\ 1 \end{array}\right)=<br />
k_1\left(\begin{array}{c}3 \\ -1 \end{array}\right)+<br />
k_2\left(\begin{array}{c}1 \\ 1 \end{array}\right)<br />

    Now just figure out what the constants are. You know how to read this matrix stuf huh? That last one is just:
    <br />
5=3k_1+k2<br /> <br />
1=-k_1+k_2<br />
    Same dif for reading the other one.

    Edit: Oh yea:

    Welcome to PF and

    "Differential Equations" by Blanchard, Devaney, and Hall is one DE book I use.
    PF?
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