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Method of undeterminant coefficeint

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

    Find a general solution of the following de:
    y" - 6y' -7y = 7t^2 - 9

    2. Relevant equations



    3. The attempt at a solution

    y^2 - 6y - 7
    y^2-7y+y-7
    y(y-7) 1(y-7)
    y = -1 or y = 7

    yc = C_1e^-t + C_2e^7t

    yp = 7At^2 + Bt + D - 9E
    y'p = 14At + B - 9
    y"p = 14A

    plug it back in:
    14A - 84At - 6B + 54 - 49At^2 -7Bt - 7D = 7t^2-9

    -84At - 7Bt = 0
    14A - 6B + 54
    A = -1/7

    Is this good so far?
     
  2. jcsd
  3. Oct 1, 2009 #2

    Mark44

    Staff: Mentor

    Your particular solution is more complicated that it needs to be. All you need is the following:
    yp = At2 + Bt + C.

    As for the particular solution you found, if it satisfies the differential equation, then you know it's the one.
     
  4. Oct 1, 2009 #3
    why is the 7 gone?
     
  5. Oct 1, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Why would there be any "7" to begin with? If "A" is some unknown coefficient then "7A" is also an unknown coefficient. There is no reason to have the "7" at all! And you don't need the "-9E" because you can treat "D- 9E" as a single uknown number as well.
     
  6. Oct 1, 2009 #5

    Mark44

    Staff: Mentor

    Because you don't need it. A is a constant and so is 7A, so why not work with the simpler one? Also D - 9E is a constant, so why not just simplify it to C and work with that instead?
     
  7. Oct 1, 2009 #6
    so is this it:

    y^2 - 6y - 7
    y^2-7y+y-7
    y(y-7) 1(y-7)
    y = -1 or y = 7

    yc = C_1e^-t + C_2e^7t

    yp = At^2 + Bt + D
    y'p = 2At + B
    y"p = 2A


    2A - 12At - 6B - 7At^2 - 7Bt - 7D

    -7A = 7
    A = -1

    (-12A - 7B) = 0
    B = 12/7

    2A - 6B - 7D = -9
    -2 - 72/7 - 7D = -9
    49/7 - 72/7 = 7D
    -23/7 = 7D
    D = -23/49
     
  8. Oct 1, 2009 #7

    Mark44

    Staff: Mentor

    You should give your particular solution (not just give the coefficients). Then check and see whether yp'' - 6yp' - 7y = 0 is a true statement.

    If so, then you have it (your complementary solution is good). The general solution is the complementary solution plus your particular solution. That's what you should present as the answer to this problem.

    BTW, you are omitting the fact that your characteristic equation is in fact an equation, meaning it needs an = in there. It's common to use some letter other than y so as not to confuse this equation with your original differential equation. A letter that is commonly used is r.

    So the characteristic equation would be
    r2 - 6r - 7 = 0
    etc.
     
  9. Oct 1, 2009 #8
    I checked and it all looked okay, the general solution should then be:

    y = -t^2 + (12/7)t -23/49 + C_1e^-t + C_2e^7t

    thanks for the suggestions
     
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