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Find a particular solution to the differential equation?

  1. Dec 13, 2011 #1
    1. The problem statement, all variables and given/known data

    Find a particular solution to the differential equation
    -6y''+5y'-1y
    =
    -1t^2+1t+1e^2t

    2. Relevant equations

    y = y(c) + y(p)

    3. The attempt at a solution

    First solve the homogeneous equation -6y" + 5y' - y = 0.
    The characteristic equation is -6r^2 + 5r - 1 = 0 --> (1-2r)(1-3r) = 0 --> r = 1/2 or r = 1/3
    So y(c) = Ae^(t/2) + Be^(t/3)
    Now using methods of undetermined coefficients we assume y(p) = Ct^2 + Dt + E + Fe^(2t)
    --> y'(p) = 2Ct + D + 2Fe^(2t)
    --> y"(p) = 2C + 4Fe^(2t)
    Substituting into the equation yields
    LHS
    = -6[2C + 4Fe^(2t)] + 5[2Ct + D + 2Fe^(2t)] - [Ct^2 + Dt + E + Fe^(2t)]
    = -12C - 24Fe^(2t) + 10Ct + 5D + 10Fe^(2t) - Ct^2 - Dt - E - Fe^(2t)
    = (-C)t^2 + (10C - D)t + (5D - E - 12C) + (-15F)e^(2t)
    Comparing the coefficients on the LHS and RHS we have
    -C = -1 --> C = 1
    10C - D = 1 --> D = 9
    5D - E - 12C = 0 --> E = 33
    -15F = 1 -- F = -1/15
    So y(p) = t^2 + 9t + 33 - (1/15)e^(2t)
    Hence y = Ae^(t/2) + Be^(t/3) + t^2 + 9t + 33 - (1/15)e^(2t)
     
  2. jcsd
  3. Dec 13, 2011 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    You will find this gets moved over to the calculus forum.

    By the way, do you have a question to ask, to accompany your working?
     
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