This is a question from a differential equations practice test. I don't understand how my professor got the answer she did - she may or may not have made an error; I got a different answer when I did it and when I ran it through a Laplace transform calculator, so I just want to be sure.
$\mathcal{L}${y}(s) = $e^{-\pi s}$$\frac{0.5}{s^2 + 0.25}$ + $\frac{s+4}{s^2 + 4s + 5}$
Note U is the step function or Heaviside function - that is how we write it in our differential equations class
Her answer was y = -cos($\frac{x}{2}$)U(x - $\pi$) + $e^{-2x}$(2sinx + cosx)
She also used trig identities: sin($\frac{x}{2}$(x - $\pi$)) = sin($\frac{x}{2}$)cos(-$\frac{\pi}{2}$) + cos($\frac{x}{2}$)sin(-$\frac{\pi}{2}$) = sin($\frac{x}{2}$(x - $\pi$)) = -cos($\frac{x}{2}$)
Note that with $f_1(t) = U(t)\sin ({1 \over 2} t)$ we have $\hat{f_1}(s) = { {1 \over 2} \over s^2 + ({1 \over 2})^2}$ and hence $s \mapsto e^{-\pi s} \hat{f_1}(s)$ corresponds to $t \mapsto U(t-\pi)f_1(t-\pi) = U(t-\pi) \sin ({1 \over 2}(t-\pi))$.
Note that if $\hat{f_2}(s) = {s+4 \over s^2+4s+5} = {s+4 \over (s+2)^2+1} = 2{1 \over (s+2)^2+1} + {s+2 \over (s+2)^2+1} $, then the corresponding function $f_2(t) = U(t) e^{-2t} (2 \sin t + \cos t)$.
Since $\sin({1 \over 2}t - {\pi \over 2}) = -\cos ({1 \over 2}t )$, the corresponding time function can be written as $t \mapsto -U(t-\pi) \cos ({1 \over 2}t ) + U(t) e^{-2t} (2 \sin t + \cos t)$.