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Homogeneous equation (third order)

  1. Oct 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Find y as a function of x if


    y'''−11y''+28y'=0 y(0)=1 y'(0)=7 y''(0)=2

    I have one attempt left on this question. Could someone verify my answer for me?

    2. Relevant equations




    3. The attempt at a solution
    (use t as lamda)
    t^3-11t^2+28t=0
    t(t-4)(t-7)=0
    t= 0, 4, 7

    y = C1 e^(4x) + C2 e^(7x) + C3

    1 = C1 (1) + C2 (1) + C3 ...(1)

    y' = 4 C1 e^(4x) + 7 C2 e^(7x)

    7 = 4 C1 + 7 C2 ... (2)

    y'' = 16 C1 e^(4x) + 49 C2 e^(7x)

    2 = 16 C1 + 49 C2 ...(3)

    Using (2) and (3) to solve for C1 and C2:

    28 = 16 C1 + 28 C2 --> (2)*2
    2 = 16 C1 + 49 C2 ---> (3)
    ------------------------------
    26 = -21 C2
    C2 = (26/-21)

    2 = 16 C1 + 49 (-26/21) ... (3)
    C1 = 47/12

    1 = -26/21 + 47/12 + C3

    C3 = 1+ 26/21 - 47/12
    C3 = 121/84

    y = (47/12)e^(4x) - (26/21) e^(7x) +121/84
     
  2. jcsd
  3. Oct 17, 2009 #2

    Mark44

    Staff: Mentor

    You've done all the hard work. Checking your solution is easy.
    See if your function satisfies the differential equation. Check that for your function, y''' - 11y'' + 28y' = 0 is a true statement.
    See if your function satisfies the initial conditions.
     
  4. Oct 17, 2009 #3
    I've submitted the answer and it is incorrect. Can anyone tell me why?
     
  5. Oct 18, 2009 #4

    Mark44

    Staff: Mentor

    Did you check your solution as I suggested in my previous post? I haven't worked the problem, so can't vouch for your solution.
     
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