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Stuck on a 2nd order linear differential equation

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data
    y'' - 2y'+ 6y = 0

    y(0) = 3
    y(5) = 7

    Find a solution y(t).


    2. Relevant equations



    3. The attempt at a solution
    I found the characteristic equation: x2 - 9x = 0, which has roots at 0 and 9.

    Therefore y(t) = C1e0x + C2e9x

    Using the initial conditions to solve this:
    3 = C1 + C2

    7 = C1 + C2(e^9)

    And solving the system of equations gives
    C1 = 3
    C2 = 4.937

    Therefore y(t) = 3 + 4.937e^9.

    But this isn't the correct answer... where did I go wrong? It seems like it should work =\
     
    Last edited: Oct 9, 2009
  2. jcsd
  3. Oct 8, 2009 #2

    Dick

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    Science Advisor
    Homework Helper

    Why is it that if you take the C1 and C2 you found, that C1+C2 isn't equal to 3?
     
  4. Oct 8, 2009 #3

    Office_Shredder

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    Staff Emeritus
    Science Advisor
    Gold Member

    You solved your system of equations incorrectly

    If C1=3, then the first equation says C2=0
     
  5. Oct 9, 2009 #4

    Mark44

    Staff: Mentor

    Some questions have already been raised about the constants, but also, your solution should be a function of t.
     
  6. Oct 9, 2009 #5
    Whoops, somehow my C2 is off by a factor of 10000. Silly mistake... thanks! And yeah, I changed the x's to t's.
     
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