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Auxiliary Eqns with Complex Roots

  1. Oct 3, 2009 #1
    1. The problem statement, all variables and given/known data
    solve the initial value problem
    y''+9y = 0, y(0) = 1, y'(0) = 1


    2. Relevant equations
    gen solution form is y(t) = C1e^At*(cosBt) + C2e^At*(sinBt)
    where A is the real number and B is the imaginary number

    3. The attempt at a solution
    i just wanted to check if im doing this right.
    i started off finding the roots,
    r^2 + 9
    r = +/- 3i
    since the real is 0, e would be raised to 0 and become one leaving me with C1cos3t + C2sin3t
     
  2. jcsd
  3. Oct 3, 2009 #2

    rock.freak667

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    Yes that is correct. Now find y(0) and y'(0)
     
  4. Oct 4, 2009 #3
    since there are double roots, should i make changes to the general solution form? everything cancels out to be zero because y'' = -9cos3t -9sin3t and y = 9cos3t + 9sin3t because it was multiplied by 9 in the original equation.
     
  5. Oct 4, 2009 #4

    rock.freak667

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    double roots would mean r=3i,3i or r=-3i,-3i. You have r=0±3i. So yes your general solution is correct.

    Also from y = 9cos3t + 9sin3t, y''≠ y'' = -9cos3t -9sin3t , recheck your algebra on that one.

    Anyway you take it y= 9cos3t +9sin3t is correct.
     
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