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Second-order IVP

  1. Sep 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Find y as a function of t if 36y''-132y'+121y=0, y(0)=5, y'(0)=4

    3. The attempt at a solution

    36y''-132y'+121y=0
    36r^2-132r+121=0
    (6r-11)^2
    So, general solution

    y(x) = C1*e^(11x/6)+C2*x*e^(11x/6)
    y'(x)=(11/6)*C1*e^(11x/6)+C2*e^(11x/6)*((6x-11)-(36/121))

    y(0)= C1=5
    y'(0)= (11/6)*5+(-36/121)*C2=4
    (-36/121)*C2=(-31/6)
    C2=(3751/216)

    So,
    y(x)=5e^((11/6)t)+(3751/216)t*e^((11/6)t)

    This is for an online homework and it's wrong, but I can't figure out where I went wrong. Can someone help me out?
     
  2. jcsd
  3. Sep 18, 2009 #2

    Office_Shredder

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    [tex] y'(x) = \frac{11}{6} C_1 e^{\frac{11x}{6}} + C_2 e^{\frac{11x}{6}} + C_2 x (\frac{11}{6} e^{ \frac{11x}{6}})[/tex]

    using the product rule on the second term I get something different than what you have... how did you get (6x-11) - 36/121?
     
  4. Sep 18, 2009 #3

    Dick

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    The derivative y'(x) is wrong. I have no idea how you got what you did. To do it correctly just use the product rule on the C2*x*e^(11x/6) part.
     
  5. Sep 18, 2009 #4
    Hmmm... I integrated it instead of derivating...haha. Well I feel stupid now...

    Thanks for the help!!
     
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