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ODE question

  1. Nov 11, 2009 #1
    Could someone please look over this question

    d^2i/dt^2 + 25i = 0

    where i(0) =15 and di/dt(0) = 0

    sketch the graph of i against t(t>0) over 2 cycles

    attempt:

    General Solution

    i=Acos(5t) + Bsin(5t)

    when i(0) = 15 is applied

    A = 15

    P.S

    ip= Acos(5t) + Bcos(5t)

    ip'= -5Asin(5t) + 5Bcos(5t)

    apply di/dt (0) = 0

    therefore 0=5B

    so B=0

    giving

    i-15cos (5t)

    where t = 1/5cos t = 1

    Am I correct in saying that this calculation has been done correctly and the graph which has to be sketched is a cosine graph peaking and troughing at 15 and -15 with a period of 1
     
  2. jcsd
  3. Nov 11, 2009 #2

    LCKurtz

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    Your solution i = 15 cos(5t) is correct. You have a homogeneous equation there, so you don't need to go through the calculation of yp for a particular solution of a non-homogeneous equation. That's why you got yp = 0.

    But the period of cos(5t) is not 1. Remember the period of cos(bt) is 2pi/b.
     
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