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!!