Use the Laplace transform to solve the following initial value problem: $$y''+4y = 3\delta(t-\pi), \quad y(0)=0, y'(0)=0$$
I solve the equation based on what I learned in class, my answer is $(3/2)H(t-\pi)\sin(2t-\pi)$.
Professor's given answer is $(3/2)H(t-\pi)\sin(2t-2\pi)$
Did I make a mistake somewhere, or did my professor mistype the number $2$ in front of $\pi$?
Assuming the initial conditions are something like $y(0) = y'(0) = 0$, we have
$$Y(s) = \frac{3}{2} e^{-\pi s} \frac{2}{s^2 + 4} \implies y(t) = \frac 3 2 H(t - \pi) \sin\big(2(t - \pi)\big)$$
where we have used the fact that $$\mathcal{L}\big(\sin(2t)\big) = \frac{2}{s^2 + 4}.$$
Hence, the professor's answer is correct.