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Second Order ODE, With Initial Conditions

  1. Oct 20, 2014 #1
    1. The problem statement, all variables and given/known data
    y'' + 4y = t2 + 6et; y(0) = 0; y'(0) = 5

    2. Relevant equations

    3. The attempt at a solution

    So, getting the general solution, we have r2 + 4 = 0, so r = +/- 2i

    So the general solution is yc = sin(2t) + cos(2t)

    I then used the method of undetermined coefficients to figure that the particular solution had a form At2 + Bt + C + Det + E = 0, and found that A = 0.25, B = 0 , C = 0, D = 1.2, E = 2

    So the overall solution is y = c1sin(2t) + c2cos(2t) + 0.25t2 + 1.2et + 2

    The derivative is y' = 2c1*cos(2t) - 2c2*sin(2t) + 0.5t + 1.2et

    So I got that 0 = 0 + c2 + 0 + 1.2 + 2, which means that c2 + 3.2 = 0, which means that c2 = -3.2

    5 = 2c1 - 0 + 0 + 1.2, which means that 2c1 = 3.8, and c1 = 1.9

    So the overall equation should be y = 1.9sin(2t) - 3.2*cos(2t) + 0.25t2 + 1.2et + 2.

    But when I enter it as an answer, it tells me it's wrong. Where did I make a mistake?
     
  2. jcsd
  3. Oct 20, 2014 #2

    SteamKing

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    No, this is only the complementary solution, i.e. the solution to the homogeneous equation.

    Why do you have two different constant terms?

    The general solution of the ODE is the sum of the particular solution and the complementary solution.

    Also, you can't apply the initial conditions solely to the complementary solution. These must be applied to the general solution of the ODE to work out the values of the undetermined coefficients.

    http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
     
  4. Oct 20, 2014 #3
    Alright, so for the particular solution:
    yp = At^2 + Bt + C + De^t
    yp' = 2At + B + De^t
    yp'' = 2A + De^t

    2A + De^t + 4(At^2 + Bt + C + De^t) = t^2 + 6e^t

    2A + De^t + 4At^2 + 4Bt + 4C + De^t = t^2 + 6e^t

    So grouping the terms together, you get 4A = 1, so A = 0.25

    De^t = 6e^t, so D = 6

    2A + 4C = 0
    2(1) + 4C = 0
    4C = -2
    C = -0.5

    So yp = 0.25t^2 - 0.5 + 6e^t

    y = c1sin(2t) + c2cos(2t) + 6e^t + 0.25t^2 - 0.5

    y' = 2c1*cos(2t) - 2c2*sin(2t) + 0.5t + 6e^t

    0 = 0 + c2 + 6 + 0 - 0.5
    -5.5 = c2

    5 = 2c1 + 6

    2c1 = -1

    c1 = -0.5

    So the final solution is y = -0.5sin(2t) -5.5 cos(2t) + 6e^t + 0.25t^2 - 0.5. Does this look correct?
     
  5. Oct 20, 2014 #4

    SteamKing

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    You can check it yourself. It's good practice.

    Does it satisfy the original ODE?

    Does it satisfy the initial values prescribed for y and y'?
     
  6. Oct 20, 2014 #5
    After some checking, I finally entered the answer y = 1.9*sin(2t) - 1.075*cos(2t) + 0.25t^2 - 0.125 + 1.2e^t. Thank you so much for your help! :)
     
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