Using Laplace transform to solve IVP involving complex roots I am attempting to solve the following IVP :
$$ y'' + y' + \frac54 y =t -(t-\pi /2 )u_{\pi /2}(t) \quad, y(0) =0 ,\quad y'(0) = 0. \tag{1}$$
My reasoning is as follows,
\begin{align*}
\mathcal{L}[y'' + y' + \frac54 y] &= \mathcal{L}[t -(t-\pi /2 )u_{\pi /2}(t)] \\
\mathcal{L}[y''] + \mathcal{L}[y'] + \frac54 \mathcal{L}[y] &= \mathcal{L}[t] - \mathcal{L}[(t-\pi /2 )u_{\pi /2}(t)]\\
&\hspace{-5cm} \text{Letting $Y(s) = \mathcal{L}[y(t)]$,}\\
Y(s)(s^{2} + s + 5/4 ) &= \frac{1}{s^2} -\frac{e^{\frac{-\pi s}{2}}}{s^2}\\
\implies Y(s) = \frac{1- e^{\frac{-\pi s}{2}}}{s^2 (s^{2} + s + 5/4 )}
\end{align*}
Since $s^{2} +s +5/4 $ has complex roots I can't do a simple partial fraction decomposition and hence must go with 
\begin{gather*}
\frac{A}{s^{2}} + \frac{Bs + C}{s^2 +s + 5/4} = \frac{1- e^{\frac{-\pi s}{2}}}{s^2 (s^2 + s +5/4)} \tag{2}\\ 
\implies A(s^2 +s + 5/4) + (Bs+C)(s^2) = 1- e^{\frac{-\pi s}{2}}
\end{gather*}
Now I will be solving for A,B,C
\begin{align*}
&s=0 \implies \frac{5A}{4} = 0 \implies A = 0. \\
&s =1 \implies \frac{5A}{4} -B + C = 1- e^{\pi / 2} \\
&\hphantom{s =1 \implies \frac{5A}{4}} \therefore C-B = 1- e^{\pi / 2}\\
&s =1 \implies \frac{13A}{4} + B + C = 1- e^{-\pi /2}\\
&\hphantom{s =1 \implies} \frac{13A}{4}+ 2B + C-B = 1-e^{-\pi/2} \\
&\hphantom{s =1 \implies} \therefore B = \frac{e^{\pi/2} - e^{-\pi/2}}{2} = \sinh (\pi/2)\\
&\hphantom{s =1 \implies} \therefore C = 1 - e^{-\pi/2} - \sinh(\pi/2)
\end{align*}
From the Laplace transformation table it is known that 
$$ \mathcal{L}[e^{-bt}\sin \omega t] = \frac{\omega}{(s+b)^{2} +\omega^{2}} \ \ \text{and} \ \ \mathcal{L}[e^{-bt}\cos(\omega t)] = \frac{s+b}{(s+b)^2 +\omega^2}, \tag{3}$$ 
hence I must complete the square in the denominator of Equation 2, yielding
$$ \frac{A}{s^2} + \frac{Bs+C}{s^2 + s + 5/4} = \frac{A}{s^2} + \frac{Bs+ C}{\left(s+\frac12 \right)^2 + 1}.$$
Replacing $B,C$ with the previously found values, 
\begin{gather}
\frac{Bs+ C}{\left(s+\frac12 \right)^2 + 1} = \frac{\sinh(\pi/2) (s) + 1 - 3^{-\pi /2} - \sinh (\pi /2)}{\left(s+\frac12 \right)^2 + 1}
\end{gather}
Now i try to express this as a linear combination of the equations in Equation 3, using $\alpha$ and $\beta$: 
\begin{gather}
\frac{\sinh(\pi/2) (s) + 1 - 3^{-\pi /2} - \sinh (\pi /2)}{\left(s+\frac12 \right)^2 + 1} = \frac{\alpha (1)}{\left(s+\frac12 \right)^2 + 1} + \frac{\beta(s+ \frac12 )}{\left(s+\frac12 \right)^2 + 1} \\
\implies \beta = \sinh (\pi/2) \\
\implies \alpha + \frac{\beta}{2} = 1 + e^{-\pi/2} - \sinh(\pi/2) \\
\therefore \alpha = 1 + e^{-\pi/2} - \frac{3}{2} \sinh(-\pi/2)
\end{gather}
Finally, 
$$ Y(s) = 1 + e^{-\pi/2} - \frac{3}{2} \sinh(-\pi/2) \left(\frac{1}{(s+\frac12 )^2 +1 }\right) +\sinh (\pi/2) \left(\frac{s+\frac12}{(s+\frac12 )^2 +1}\right) $$
Since $y(t) = \mathcal{L}^{-1} [Y(s)]$, 
$$ y(t) = (1 + e^{-\pi/2} - \frac{3}{2} \sinh(-\pi/2)) e^{-t/2}\sin(t) + (\sinh(\pi/2)) e^{-t/2}\cos(t).$$
All of this seem quite long and I'm not entirely sure if my reasoning is correct, could someone give me a hint or confirm ? 

$$f(s)= \dfrac 1 {s^2(s^2+s+5/4)}$$
Should be decomposed as 
$$f(s)= \dfrac A s + \dfrac B{s^2} +\dfrac {Cs+D} {s^2+s+5/4}$$
And both $A,B$ aren't zero
