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Differential Equation Involving Heaviside function

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data
    y''+y= H(3pi)(t)

    y(0)=1

    y'(0)=0

    H(3pi)(t) means the heaviside function such that it is 1 above 3pi and 0 otherwise.


    2. Relevant equations



    3. The attempt at a solution

    Taking the laplace of both sides, I have that:

    L(y) = (L(H(3pi)(t))+s) / (s^2 + 1)

    The laplace transform of H(3pi)(t) is e^(-3pi(s))/s, so plugging that in and doing algebra I get:

    e^(-3pis)(1/(s(s^2+1)) + s/(s^2+1)

    For the first, I split into partial fractions. The second is ready to be inverted already.

    e^(-3pis)((1/s) - s/(s^2+1)) + s/(s^2 + 1)

    I find the inverse laplace of this to be:


    H(3pi)(t)(1-cos(t-3pi)) + cos(t)

    or

    H(3pi)(t)(1+cos(t)) + cos(t)

    Wolfram has a coefficient of 2 on the conditional cosine. How do I get that?


    Thanks again!!
     
  2. jcsd
  3. May 1, 2013 #2

    Mark44

    Staff: Mentor

    Your work looks OK to me. Without seeing what you entered into Wolfram it's hard to say where they got that 2 you mention.

    Your notation for the Heaviside function is a bit cumbersome. Instead of H(3pi)(t), which suggests the product of H(3pi) and t, I would write it as H(t - 3##\pi##).
     
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