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Math Help - Ivp

  1. #1
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    Ivp

    Given the IVP y'' - 4y = 0, y(0) = 2, y'(0) = \alpha

    1.) Find sol'n of the IVP in terms of \alpha

    2.) Find \alpha such that as t \rightarrow \infty the sol'n approached zero.
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  2. #2
    Eater of Worlds
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    First, we find the auxiliary equation by the quadratic.

    m^{2}-4=(x+2)(x-2)=0

    This gives us:

    C_{1}e^{2x}+C_{2}e^{-2x}

    Now, by using y(0)=2, we can find C1:

    C_{1}e^{2(0)}+C_{2}e^{-2(0)}=2

    C_{1}=2-C_{2}

    But y'=2C_{1}e^{2x}-2C_{2}e^{-2x}

    Then using the other IC, we can find C2:

    2(2-C_{2})e^{2(0)}-2C_{2}e^{-2(0)}=a

    C_{2}=1-\frac{1}{4}a

    Then C_{1}=\frac{1}{4}a+1

    Therefore, the IVP in terms of a is:

    (1+\frac{1}{4}a)e^{2x}+(1-\frac{1}{4}a)e^{-2x}

    Now, find the limit for the last portion of the problem. It should be kind of easy to see what a is now.

    \lim_{x\rightarrow{\infty}}[(1+\frac{a}{4})e^{2x}]+\lim_{x\rightarrow{\infty}}[(1-\frac{a}{4})e^{-2x}]

    What is the limit on the right when x is unbounded?. What will give 0?
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  3. #3
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    Regarding the limit... won't this only be 0 when alpha (or a in your case) = -4? That'll take care of the first part and the 2nd part
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