Solve the initial value problem via convolution of Laplace transforms 
$y''-y=\sin2t+1,  y(0) = 1, y'(0) = -1$

Hey all, I am not sure if I am completing all the steps correctly to solve this problem via convolution. I am pretty sure I am not doing it correctly, but am not sure what is going wrong. 
First I apply Laplace transforms to  everything. $$\mathcal{L}[y''] - \mathcal{L}[y]  = \mathcal{L}[\sin2t] + \mathcal{L}[1]$$
Plugging in initial values and doing algebra gives me $$Y(s) = \left(\frac{1}{s^2+4} + \frac{1}{s} \right) \left(\frac{1}{s+1} \right)$$
Now I let 
\begin{align*}
\mathcal{L}[f](s) &= \left(\frac{1}{s^2+4} + \frac{1}{s} \right) \\
\mathcal{L}[g](s) &= \left(\frac{1}{s+1} \right)
\end{align*}
I get $f(t) = \sin2t+1$ and $g(t) = e^{-t}$ via inverse laplace transform.
Now that I have $f$ and $g$, I want to do a convolution $f*g = \int_{0}^{t} f(x)g(t-x)dx$ , but I am unable to solve the integral so I think I have done something wrong before getting to this step.
 A: $$y''-y=\sin2t+1,  y(0) = 1, y'(0) = -1$$
I got this:
$$Y(s) = \left(\frac{2}{s^2+4} + \frac{1}{s} \right) \left(\frac{1}{s^2-1} \right)+\dfrac 1 {s+1}$$
Which is decomposed as:
$$Y(s) = -\frac{2}{5(s^2+4)} + \frac 1{5(s-1)} +\frac{s}{s^2-1} - \frac{1}{s}  +\dfrac 4 {5(s+1)}$$
$$y(t)=-1-\frac 15 \sin(2t)+\frac 15e^t+\cosh(t)+\frac 45e^{-t}$$
Finally:
$$\boxed {y(t)=-1-\frac 15 \sin(2t)+\frac {7}{10}e^t+\frac {13}{10}e^{-t}}$$

Edit:
I didn't see that OP need to use Convolution Integral;
$$Y(s) = \left(\frac{2}{s^2+4} + \frac{1}{s} \right) \left(\frac{1}{s^2-1} \right)+\dfrac 1 {s+1}$$
$$y(t) = \int_0^t \sinh (\tau)\sin(2(t-\tau))d\tau +\int_0^t \sinh (t-\tau)d\tau +e^{-t}$$
$$y(t) = \int_0^t \sinh (\tau)\sin(2(t-\tau))d\tau +\cosh(t)-1 +e^{-t}$$
Integrate by part for the first integral.
$$I=\int_0^t \sinh (\tau)\sin(2(t-\tau))d\tau$$
$$I=\frac 25\sinh t -\frac 15\sin(2 t)$$
Finally :
$$ y(t) = \frac 25\sinh t -\frac 15\sin (2t) +\cosh(t)-1 +e^{-t}$$
$$\boxed {y(t)=-1-\frac 15 \sin(2t)+\frac {7}{10}e^t+\frac {13}{10}e^{-t}}$$
A: You computed your initial Laplace transform incorrectly.  Notice that 
\begin{eqnarray*}
\mathcal{L}[y''] - \mathcal{L}[y] & = & s^2\mathcal{L}[y] - y(0) - y'(0) - \mathcal{L}[y] \\
& = & (s^2-1)\mathcal{L}[y] - 1 - (-1) \;\; =\;\; (s^2-1)\mathcal{L}[y].
\end{eqnarray*}
You divided by the wrong term on the right hand side.  You should've obtained
$$
Y(s) \;\; =\;\; \left (\frac{1}{s^2+4} + \frac{1}{s}\right )\left (\frac{1}{s^2 - 1}\right )
$$
hence $f(t) = \frac{1}{2}\sin(2t) + 1$.  Because $\frac{1}{s^2-1} = \frac{1}{(s+1)(s-1)} = \frac{1}{2(s+1)} - \frac{1}{2(s-1)}$, then we find that $g(t) = \frac{1}{2}e^{-t} - \frac{1}{2}e^t$.
