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Undetermined Coefficients Initial Value Question

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data

    y''-4y=60e4t

    y(0)=9 and y'(0)=2

    2. Relevant equations

    none in particular

    3. The attempt at a solution

    First I solved for the auxiliary equation.

    r2-4r
    r(r-4)
    r=0, 4

    So the general solution for the homogenous form is C1e0x+C2e4x where C1 and C2 are unknown coefficients to be found.

    The particular solution is calculating by considering Yp= AXe4x
    Differentiating that twice, and solving for Yp, I get Yp=15xe4x

    So the general solution for this is obtained by adding the homogeneous and the particular. So I get: C1e0x+C2e4x +15xe4x

    I differentiated this equation and got 4C2+15e4x+60x4x

    So then I got 4C2+15=2, so C2=-3.25
    Also, from the general solution C1+C2=9, so C1=12.25

    The final answer I got was (12.25)e0x+-3.25e4x+15xe4x but this is wrong. Any ideas?
     
  2. jcsd
  3. Mar 10, 2010 #2
    you are not using the method of undeteremined coefficients i don't think. try searching that up on google.
     
  4. Mar 10, 2010 #3

    vela

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    Your characteristic equation is wrong.

     
  5. Mar 10, 2010 #4
    OHHH I see...thanks for pointing that out.. so the equation would be (r+2)(r-2)....
     
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