Determining initial conditions for 2nd order non-homogenous differential equation With the help of wxMaxima (math not being my language), I am trying to find out the impulse response for a 2nd order transfer function:
$$H(s)=\frac{a_2s^2+a_1s+a_0}{b_2s^2+b_1s+b_0}=\frac{2.3s^2+0.1s+4.5}{1.3s^2+0.7s+1.1}$$
If I use the inverse Laplace, ilt(partfrac(H(s), s), s, t);, it gives the correct response:
$$\exp\left(-\frac{7 t}{26}\right)\left( \frac{9668 \sin{\left( \frac{\sqrt{523} t}{26}\right) }}{169 \sqrt{523}}-\frac{148 \cos{\left( \frac{\sqrt{523} t}{26}\right) }}{169}\right)$$
(I omitted the ilt(a2/b2,s,t) term since it cannot be plotted) but if I am trying to implement H(s) into a differential equation:
$$b_2y''(t)+b_1y'(t)+b_0y(t)=a_2x''(t)+a_1x'(t)+a_0x(t)$$
$$1.3y''(t)+0.7y'(t)+1.1y(t)=2.3x''(t)+0.1x'(t)+4.5x(t)$$
and use wxMaxima's builtin ode2() and ic2() to solve it:
b2*'diff(y, t, 2) + b1*'diff(y, t) + b0*y = a2*'diff(x, t, 2) + a1*'diff(x, t) + a0*x;
ode2( ''%, y, t);
ic2(%th(1), t=0, y=0, 'diff(y,t)=1);

the result is different. I used for the initial conditions $y'$=1, since the input is the Dirac function and I cannot make it $\infty$, and $y$=0, and the result is this:
$$\exp\left(-\frac{7 t}{26}\right)\left( -\frac{\sin{\left( \frac{\sqrt{523} t}{26}\right) } \left( 315 \sqrt{523} x-279 \sqrt{523}\right) }{5753}-\frac{\cos{\left( \frac{\sqrt{523} t}{26}\right) } \left( 45 x+1\right) }{11}\right) +\frac{45 x+1}{11}$$

The red trace is the inverce Laplace. Both traces are missing the initial Dirac because it can't be plotted. For the same reason, the ODE is plotted with x=0, and also since any sort of impulse (e.g. if t<0.1 then 1 else 0) or similar don't work.
If I use an initial output value for $y$ of -148/169 (which is the cos() term of the inverse Laplace) and I further tweak the $y'$ initial condition to be 2.33 (with the hammer), I get this:

It looks like there is a DC component that just stays there, no matter what. Since my intuition fails, maybe the hammer won't, so I thought of differentiating the answer with changed initial conditions, which would give zero steady-state. Here's with $y$=-2.22 and $y'$=-148/169:

but this is some brute-force that I don't think it applies.
My questions: Is what I am doing the correct way of solving the ODE? If no, how? If yes, how should I determine the correct initial conditions?
 A: A more systematic way considers a solution $y_0$ of $$L(y_0)=b_2y_0''+b_1y_0'+b_0y_0=x,$$ $y(t)=0$ for $t<0$. Then \begin{align}
y&=a_0y_0+a_1y_0'+a_2y_0''\\
&=\left(a_0-\frac{a_2b_0}{b_2}\right)y_0+\left(a_1-\frac{a_2b_1}{b_2}\right)y_0'+ \frac{a_2}{b_2}x
\end{align}
is a solution for the given ODE with derivative
\begin{align}
y'&=\left(a_0-\frac{a_2b_0}{b_2}\right)y_0'+\left(a_1-\frac{a_2b_1}{b_2}\right)y_0''+ \frac{a_2}{b_2}x'\\
&=-\frac{b_0}{b_2}\left(a_1-\frac{a_2b_1}{b_2}\right)y_0+\left(a_0-\frac{a_2b_0+a_1b_1}{b_2}+\frac{a_2b_1^2}{b_2^2}\right)y_0'+\left(\frac{a_1}{b_2}-\frac{a_2b_1}{b_2^2}\right)x+\frac{a_2}{b_2}x'
\end{align}

For $x=\delta$ we get $y_0=uv$ where $u(t)=0$ for $t<0$, $u(t)=1$ for $t>0$ is the unit jump and $v$ a solution to $L(v)=0$ with $v(0)=0$, $v'(0)=\frac1{b_2}$, which implies $y_0'=δv+uv'=δv(0)+uv'=uv'$. Thus 
\begin{alignat}{1}
y(0_{+0})&=\left(a_1-\frac{a_2b_1}{b_2}\right)v'(0)&=\frac{a_1}{b_2}-\frac{a_2b_1}{b_2^2}
\\
y(0_{+0})&=\left(a_0-\frac{a_2b_0+a_1b_1}{b_2}+\frac{a_2b_1^2}{b_2^2}\right)v'(0)&=\frac{a_0}{b_2}-\frac{a_2b_0+a_1b_1}{b_2^2}+\frac{a_2b_1^2}{b_2^3}
\end{alignat}
