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Math Help - 2nd order ODE with a catch

  1. #1
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    2nd order ODE with a catch

    I have an odd ODE question that I could use a hand with...

    ----------------

    y'' - y' - 2y = 0

    y(0) = a

    y'(0) = 2

    Find "a" so that the solution approaches 0 as t approaches infinity.

    ----------------


    So here's what I've done:

    r^2 - r - 2 = 0

    r = -1, 2

    y = C_{1}e^{-t} + C_{2}e^{2t}

    a = C_{1} + C_{2}

    C_{2} = a - C_{1}

    y' = -C_{1}e^{-t} + 2C_{2}e^{2t}

    2 = -C_{1} + 2C_{2}

    C_{1} = 2C_{2} - 2

    C_{1} = 2a - 2C_{1} - 2

    C_{1} = \frac{2a - 2}{3}


    y = \frac{C_{1}}{e^t} + C_{2}e^{2t}


    So, unless I've made a mistake somewhere, I've done most of the work. But I don't understand how I can get an "a" that will cause the solution to approach 0 as t approaches infinity. Since the second term is not part of a fraction, I'm not sure how this can be done.

    Can someone help me to understand what I'm missing?
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  2. #2
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    Re: 2nd order ODE with a catch

    Quote Originally Posted by Lancet View Post
    I have an odd ODE question that I could use a hand with...

    ----------------

    y'' - y' - 2y = 0

    y(0) = a

    y'(0) = 2

    Find "a" so that the solution approaches 0 as t approaches infinity.

    ----------------


    So here's what I've done:

    r^2 - r - 2 = 0

    r = -1, 2

    y = C_{1}e^{-t} + C_{2}e^{2t}

    a = C_{1} + C_{2}

    C_{2} = a - C_{1}

    y' = -C_{1}e^{-t} + 2C_{2}e^{2t}

    2 = -C_{1} + 2C_{2}

    C_{1} = 2C_{2} - 2

    C_{1} = 2a - 2C_{1} - 2

    C_{1} = \frac{2a - 2}{3}


    y = \frac{C_{1}}{e^t} + C_{2}e^{2t}


    So, unless I've made a mistake somewhere, I've done most of the work. But I don't understand how I can get an "a" that will cause the solution to approach 0 as t approaches infinity. Since the second term is not part of a fraction, I'm not sure how this can be done.

    Can someone help me to understand what I'm missing?
    If y --> 0 as t --> +oo then C2 has to equal zero.

    So you have y = C1 e^(-t) and since y(0) = a, C1 = a.

    So y = a e^(-t).

    Now use y'(0) = 2 to get a.
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  3. #3
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    Re: 2nd order ODE with a catch

    Quote Originally Posted by mr fantastic View Post
    If y --> 0 as t --> +oo then C2 has to equal zero.

    So you have y = C1 e^(-t) and since y(0) = a, C1 = a.

    So y = a e^(-t).

    Now use y'(0) = 2 to get a.
    Thank you so much for that - I just couldn't see what I was missing. In retrospect, it seems obvious, and worth kicking myself over.

    Thanks again!
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  4. #4
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    Re: 2nd order ODE with a catch

    No, no, don't kick yourself. Let us do it!
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