H(3pi)(t) means the heaviside function such that it is 1 above 3pi and 0 otherwise.
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)
H(3pi)(t)(1+cos(t)) + cos(t)
Wolfram has a coefficient of 2 on the conditional cosine. How do I get that?