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Math Help - Find the solution given the initial value problem

  1. #1
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    Find the solution given the initial value problem

    Also determine its behavior as t increases.

    6y'' - 5y' + y = 0 y(0) = 4 y'(0) = 0

    I found the general solution to be y = C1*e^(t/2) + C2*e^(t/3)

    C1 + C2 = 4
    (1/2)*C1 + (1/3)*C2 = 0

    I substituted in C1 and solved that way getting the 4 for C1 and 3 for C2. Obviously I did something wrong here, any ideas? Thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by pakman View Post
    Also determine its behavior as t increases.

    6y'' - 5y' + y = 0 y(0) = 4 y'(0) = 0

    I found the general solution to be y = C1*e^(t/2) + C2*e^(t/3)

    C1 + C2 = 4
    (1/2)*C1 + (1/3)*C2 = 0

    I substituted in C1 and solved that way getting the 4 for C1 and 3 for C2. Obviously I did something wrong here, any ideas? Thanks!
    Let y = e^mt
    Then the characteristic equation is
    6m^2 - 5m + 1 = 0
    => (3m - 1)(2m - 1) = 0
    => m = 1/3 or m =

    So y = Ae^(t/3) + Be^(t/2)
    => y’ = (1/3)Ae^(t/3) + (1/2)Be^(t/2)

    y(0) = 4
    => y(0) = 4 = Ae^0 + Be^0 = A + B
    So A + B = 4 …………………(1)

    y’(0) = 0
    => y’(0) = 0 = (1/3)Ae^0 + (1/2)Be^0 = (1/2)A + (1/2)B
    So (1/3)A + (1/2)B = 0 ………...(2)

    So we have the system:
    A + B = 4 ………………..…(1)
    (1/3)A + (1/2)B = 0 ………...(2)

    A + B = 4 ………………..…(1)
    A + (3/2)B = 0 ……………...(3) = (2) * 3

    => (1/2)B = -4
    => B = -8

    But A + B = 4
    => A – 8 = 4
    => A = 12

    So y = 12e^(t/3) – 8e^(t/2)
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  3. #3
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    Thanks! If I used the quadratic to find the characteristic, how would I know which one is A or B?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by pakman View Post
    Thanks! If I used the quadratic to find the characteristic, how would I know which one is A or B?
    we decided at the beginnig. i used A to be the coefficient of e^(t/3) and B to be the coefficient of e^(t/2), so you choose where you want which constant from the beginning, they are just arbitrary constants
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