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Math Help - Help :(

  1. #1
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    Help :(

    Okay.. Can anyone explain me, how to solve this problem:

    y''+4y'+4=0.. when y(0)=2 and y'(0)=0
    Last edited by mr fantastic; September 25th 2009 at 05:54 PM. Reason: Made the boundary conditions clearer.
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  2. #2
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    Quote Originally Posted by MissWonder View Post
    Okay.. Can anyone explain me, how to solve this problem:

    y''+4y'+4=0.. when y(0)=2 and y'(0)=0
    The Characteristic Equation is

    m^2 + 4m + 4 = 0

    (m + 2)^2 = 0

    m = -2.


    Since the solution of the characteristic equation is repeated, the solution will be of the form

    y(x) = C_1e^{-2x} + C_2xe^{-2x}.

    We can also see that

    y'(x) = -2C_1e^{-2x} + C_2e^{-2x} -2C_2xe^{-2x}.


    Applying the initial conditions gives

    2 = C_1 + C_2 and 0 = -2C_1 + C_2.


    Solving the equations simultaneously gives

    2 = 3C_1

    \frac{2}{3} = C_1


    2 = \frac{2}{3} + C_2

    \frac{4}{3} = C_2.



    Thus y(x) = \frac{2}{3}e^{-2x} + \frac{4}{3}xe^{-2x}.
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  3. #3
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    hmm

    Applying the initial conditions gives

    and .

    Where do you get that last part from?
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  4. #4
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    Quote Originally Posted by MissWonder View Post
    Applying the initial conditions gives

    and .

    Where do you get that last part from?
    The reply given was quite clear. Go back and read it carefully. Especially the part where the given initial condition y'(0) = 0 is substituted into the rule for y'.
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