I am trying to understand the relation between the solution of differential equation in Laplace space and matrix inverse/pseudo-inverse problems.
Consider the system of ODEs:
$$\dot{\mathbf{x}}(t) = \mathcal{L}\mathbf{x}(t)$$
where $\mathbf{x}(t)$ is a complex vectors, and $\mathcal{L}$ is a complex singular matrix.
Now I can transform to Laplace space and rewrite this equation as a simple linear problem:
$$(s\mathbb{1} - \mathcal{L})\tilde{\mathbf{x}}(s) = \mathbf{x}_0 $$
$\mathbf{x}_0$ is the initial condition of the ODE. For my specific purposes I can restrict myself to purely imaginary values of the Laplace variable, i.e. $s = i\omega$. For all values of $\omega$ for which the matrix is $(i\omega\mathbb{1}-\mathcal{L})$ is non-singular, I can write the solution as a matrix inverse:
$$\tilde{\mathbf{x}}(\omega) = (i\omega\mathbb{1}-\mathcal{L})^{-1}\mathbf{x}_0$$
Clearly this does not hold in situation where the matrix is singular, e.g. $\omega=0$.
If we take this point as an example, I thought that I could use the Pseudo-inverse in place of the inverse to solve $\mathcal{L}\tilde{\mathbf{x}}(\omega=0) = -\mathbf{x}_0$. This approach does sometimes work, but in a few cases comparing this to the time domain solution I find: $$ \tilde{\mathbf{x}}(\omega=0) = \int\limits_0^\infty \mathbf{x}(t)dt \neq -\mathcal{L}^{+}\mathbf{x}_0 $$ where $\mathcal{L}^{+}$ is the pseudo inverse of $\mathcal{L}$.
So I guess my question is two fold: given that the pseudo-inverse is unique, why does this approach not work? Is there a way to modify the pseudo inverse approach to get the correct answer?