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Math Help - Second Order DE

  1. #1
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    Second Order DE

    y = c1e4x + c2e-x is the general solution of the differential equation y'' - 3y' - 4y = 0 on (-infinity, infinity). Find a member of the family that is a solution with initial conditions y(0) = 1 and y'(0) = 2.
    What is the value of c1:
    What is the value of c2:

    So...a=1, b=3and c=-4
    using the quad equation.... 4+/-Sqrt16-16/2a

    The discriminat equals 0, so will use -b/2a for m1 and m2

    So where do I go from here to find c1 and c2? We briefly touched this in class and I was trying to figure it out myself.
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  2. #2
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    Quote Originally Posted by latavee View Post
    y = c1e4x + c2e-x is the general solution of the differential equation y'' - 3y' - 4y = 0 on (-infinity, infinity). Find a member of the family that is a solution with initial conditions y(0) = 1 and y'(0) = 2.
    What is the value of c1:
    What is the value of c2:

    So...a=1, b=3and c=-4
    using the quad equation.... 4+/-Sqrt16-16/2a

    The discriminat equals 0, so will use -b/2a for m1 and m2

    So where do I go from here to find c1 and c2? We briefly touched this in class and I was trying to figure it out myself.
    If y''(x) - 3y'(x) - 4y(x) = 0, the characteristic equation is

    m^2 - 3m - 4 = 0

    (m - 4)(m + 1) = 0

    m = 4 or m = -1.


    Thus the solution is of the form

    y(x) = C_1e^{4x} + C_2e^{-x}.


    You should also be able to see that

    y'(x) = 4C_1e^{4x} - C_2e^{-x}.



    Using your initial conditions y(0) = 1 and y'(0) = 2, we get

    1 = C_1 + C_2 and

    2 = 4C_1 - C_2.


    Solving these equations simultaneously gives

    3 = 5C_1

    C_1 = \frac{3}{5}.


    1 = \frac{3}{5} + C_2

    C_2 = \frac{2}{5}.



    Therefore, the solution is

    y(x) = \frac{3}{5}e^{4x} - \frac{2}{5}e^{-x}.
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  3. #3
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    Thank you

    I totally had that characteristic equation not in my head! This makes sense now! Thanks!
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