Inhomogeneous 2nd-order linear differential equation I need to solve this:
$y'' + ay' + by = x(t)$
where nothing about the form of $x$ is known, except that it is bounded and non-negative.  In addition it is known that $y(0) = 0$ and $y'(0) = 0$ (and $y''(0) = 0$ as well). I plugged this into the Wolfram ODE solver and got back a 'possible Lagrangian', which I really don't know how to use.  (I have an undergraduate degree in math but I never took differential equations.)  I also found this resource pt1, pt2, but it assumes that the solution is exponential in nature, which seems to violate my initial conditions since exponentials are never 0.  I'm hoping there's a way to get my head around this without putting myself through the full course of diff eq so any help would be much appreciated.
Update: So actually Wolfram Alpha will solve this equation (simplified slightly in this edit), it was just the 'widget' that stopped short.  However, it, as well as the techniques given in the references, explicitly assumes that the solution has an exponential form.  Apparently this is a standard assumption but it is not the case for my data.  $y$ is roughly sigmoidal.  Coding up Wolfram's solution and choosing arbitrary constants $c_1$, $c_2$ so as to minimize the error in a case where $y$ was known produced a clearly wrong answer.
So, I will sharpen my question: How to solve the above differential question without assuming that $y$ is proportional to $e^{\lambda t}$?

Let $f_\pm (t) := \exp(\lambda_\pm t)$ where $\lambda_\pm := (-a \pm \sqrt{a^2 - 4 b})/2$ be the fundamental solutions to the homogenous equation. Then the solution to the ODE, subject to $y(0)=r_0$ and $y^{'}(0)= \mu_0$, reads:
\begin{equation}
y(t) = \left(\begin{array}{cc} f_+(t) & f_-(t) \end{array}\right) \cdot
\left( \begin{array}{cc} f_+(0) & f_-(0) \\ f_+^{'}(0) & f_-^{'}(0) \end{array} \right)^{-1} \cdot
\left( \begin{array}{c} r_0 \\ \mu_0 \end{array} \right)
+
\int\limits_0^t \left(\frac{f_+(\xi) f_-(t)-f_-(\xi) f_+(t)}{f_+(\xi) f_-^{'}(\xi) - f_-(\xi) f_+^{'}(\xi)}\right) x(\xi) d\xi
\end{equation}
This can be proven using the Green's function method, for instance.
 Now, the first term on the right hand side is the generic solution to the homogenous equation and the second term is a specific solution to the inhomogenous equation.The quantity in the denominator in the integrand is called the Wronskian $W(\xi)$ and it reads:
\begin{equation}
W(\xi) = (\lambda_- - \lambda_+) e^{(\lambda_- + \lambda_+) \xi} = - \sqrt{a^2 - 4 b} e^{-a \xi}
\end{equation}
 In our case $r_0=0$ and $\mu_0 = 0$ and so the solution reads:
\begin{equation}
y(t) = \frac{-1}{\sqrt{a^2 - 4 b}} 
\left( 
e^{\lambda_- t} \int\limits_0^t e^{(\lambda_++a) \xi} x(\xi) d \xi
-
e^{\lambda_+ t} \int\limits_0^t e^{(\lambda_-+a) \xi} x(\xi) d \xi
\right)
\end{equation}
