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Is this 2nd Differential Equation Answer Correct?

  1. Nov 20, 2009 #1
    1. The problem statement, all variables and given/known data
    y'' + 9y = 0 where y(pi/3) = y'(pi/3) = 3


    2. Relevant equations

    r^2 + 9 = 0
    r^2 = -9
    r = 3i

    3. The attempt at a solution

    y = c1 e^x cos (3x) + c2 e^x sin(3x)
    y' = c1 e^x cos (3x) - 3c1 e^x sin (3x) + c2 e^x sin(3x) + 3c2 e^x cos(3x)

    y = -c1 e^(pi/3) = 3 => c1 = -3e^(-pi/3)
    y' = -c1 e^(pi/3) - 3c2 e^(pi/3) = 3
    y' = -(-3e^(-pi/3))e^(pi/3) - 3c2 e^(pi/3) = 3
    y' = 3 - 3c2e^(pi/3) = 3 ====> c2 = 0

    so
    y= -3e^(-pi/3) e^x cos(3x) + 0
     
    Last edited: Nov 20, 2009
  2. jcsd
  3. Nov 20, 2009 #2

    tiny-tim

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    Hi jonnejon! :smile:

    (have a pi: π and try using the X2 and X2 tags just above the Reply box :wink:)
    Noooo :redface: … r = ±3i gives you Ae3ix + Be-3ix (or Acos3x + Bsin3x).

    (Your solution would be for r = 1 ± 3i. :wink:)
     
  4. Nov 20, 2009 #3
    So,

    y = A cos(3x) + B sin(3x)
    y' = -3A sin(3x) + 3B sin(3x)

    y = -A = 3 => A = -3
    y'= -3B = 3 => B = -1

    So the solution is: y= 3cos(3x) - sin(3x)??
     
  5. Nov 20, 2009 #4

    tiny-tim

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    Yup! :smile:

    (except you left out the - in -3 :wink:)
     
  6. Nov 20, 2009 #5

    Mark44

    Staff: Mentor

    For y' it should by y' = -3A sin(3x) + 3B cos(3x)
    If I can jump in here, the next two lines are where you're using your initial conditions to solve for A and B in the equations above.
    What you should say in the first equation is that since y(pi/3) = 3, then
    Acos(pi) + Bsin(pi) = 3, so A*(-1) = 3, or A = -3

    Since y'(pi/3) = 3, and A = -3,
    9sin(pi) + 3B cos(pi) = 3, so 3B(-1) = 3, so B = -1

    As tiny-tim already pointed out, y = -3cos(3x) - sin(3x)
     
    Last edited: Nov 21, 2009
  7. Nov 21, 2009 #6

    tiny-tim

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    Hi Compuchip! :smile:

    Didn't know you were there! :biggrin:
     
  8. Nov 21, 2009 #7

    Mark44

    Staff: Mentor

    Sorry about that, tiny-tim.
     
  9. Nov 21, 2009 #8

    tiny-tim

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    We look alike in the dark! :biggrin:
     
  10. Nov 21, 2009 #9

    Mark44

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    Well, it's dark here, so that's good enough for me.
     
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