I am trying to solve an equation of an underdamped harmonic oscillator with a damping, and I get a weird boundary condition that perplexes me. Let me precise the issue, the equation is :
\begin{equation}
 \ddot{y} + 2\gamma \omega\_0 \dot{y} + \omega^2\_0 y = f(t) \label{eq:harmonic\_osc}
\end{equation}
And I am also interested in the case where the force $f(t)$ is an impulse as following:
$$ f(t) = A \delta(t)$$

By the applying the Laplace transform to the both sides of the equation, we get:

\begin{equation}
 s^2 y(s) - \dot{y}(t = 0 )+ 2\gamma \omega\_0 s y(s) + \omega^2\_0 y(s) = f(s)
\end{equation}
where $y(t = 0 ) = 0 $, thanks to the boundary condition that we impose. As for the $\dot{y}(t=0)$ we can get it from the integrating the harmonic oscillator equation:

\begin{equation}
 \int \limits\_{-t\_2}^t \, dt \left(\ddot{y} + 2\gamma \omega\_0 \dot{y} + \omega^2\_0 y \right)= \int \limits\_{-t\_2}^t \, dt \, f(t), 
\end{equation}
where all the values for time $-t\_2$ are equal to zero by the definition of the Laplace transform and, thus we get that $\dot{y}(t=0) = A \int \limits\_{-t\_2}^0 \, dt \, \delta(t) = A/2$

so:

$$ y(s) = \frac{A}{s^2 + 2 \gamma \omega\_0 s + \omega^2\_0} $$

Next we can calculate the inverse Laplace transform:
$$ y(t) = \frac{1}{2 \pi i }\int \limits^{c+i\infty}\_{c-i\infty} y(s) e^{st} \,ds$$

This integral is being calculated by finding residues and etc, so I get that :

$$ y(t) = \frac{A e^{-\gamma \omega\_0 t}}{\tilde{\omega}\_{0} } \sin{\tilde{\omega}\_{0} t}$$
where $\tilde{\omega}\_0 = \omega\_0 \sqrt{1-\gamma^2}$

Now we see that $y(0) = 0$, however as for the first derivative of $y(t)$ we have:

$$ \dot{y} (t) = \frac{A} {\tilde{\omega}\_0} (-\gamma \omega\_0 e^{-\gamma \omega\_0 t} \sin{\tilde{\omega}\_{0} t} + e^{-\gamma \omega\_0 t} \cos{\tilde{\omega}\_{0} t}) $$
and the $\dot{y}(0) = A,$ which is different from the initial assumption where $\dot{y}(0) = A/2$, could someone please tell me where is a catch? It seems that I have checked a lot of things but I could not find a problem.