Finding the zero-state output The input and output of a stable network are related via the following equation.
$$\frac{d^2y(t)}{d(t)} + \frac{2*dy(t)}{d(t)} + 10y(t) = \frac{dx(t)}{d(t)} + x(t)$$ 
x(t) = input, y(t) = output, u(t) = unit function.
The input is $$\frac{3u(t)}{e^t}$$
I want to find the zero-state output.
Now I have the transfer function as $$\frac{iw + 1}{-w^2 +2(iw) + 10}$$
But I'm not quite sure where to proceed from here. My intuition is to move the transfer function to the time domain through fourier transform, but I'm not sure how I would use that to continue the problem. 
 A: First find the solutions of $y''(t) + 2y'(t) + 10y = 0$. Passing to the characteristic polynomial, it has roots $\lambda_{1,2} = -1\pm 3i$, so every function of the form
$$
y_0(t) = Ae^{-t}e^{i3t} + Be^{-t}e^{-i3t}
$$
solves the homogenous equation.
Now consider the non-homogenous one. If $u$ is differentiable you get
$$
x'(t) + x(t) = 3u'(t)e^{-t}-3u(t)e^{-t}+3u(t)e^{-t} = 3u'(t)e^{-t}.
$$
I seek for particular solutions $y(t) = a(t)e^{-t}$. So $y'(t) = a'(t)e^{-t}-a(t)e^{-t}$, $y''(t) = a''(t)e^{-t}-2a'(t)e^{-t} + a(t)e^{-t}$. Substituting into the equation you get
$$
a''(t) + 9a(t) = 3u'(t).
$$
If you know $u(t)$ there is some chance to solve it, and your solution will be
$$
y_0(t) + a(t)e^{-t}
$$
for some opportune constants $A,B$.
A: $$y''(t)+2y'(t)+10y(t)=x'(t)+x(t)$$
The Fourier transform of $y(t)$ will be denoted $\quad F_t[y(t)](\omega) = Y(\omega)\quad$ on a more compact writing.
$\quad F_t[y'(t)](\omega) = -i\omega Y(\omega)$
$\quad F_t[y''(t)](\omega) = -\omega^2Y(\omega)$
$\quad F_t[x(t)](\omega) =F_t[u(t)e^{-t}](\omega) = \frac{1}{\sqrt{2\pi}(1-i\omega) }$
$\quad F_t[x'(t)](\omega) = F_t[\frac{d}{dt}\left(u(t)e^{-t}\right)](\omega) =-i\omega F_t[u(t)e^{-t}](\omega) = \frac{-i\omega}{\sqrt{2\pi}(1-i\omega) }$
$$-\omega^2Y(\omega)-2i\omega Y(\omega)+10Y(\omega)=\frac{-i\omega}{\sqrt{2\pi}(1-i\omega) }+\frac{1}{\sqrt{2\pi}(1-i\omega) } = \frac{1}{\sqrt{2\pi}}$$
$$Y(\omega)=\frac{1}{\sqrt{2\pi}\left(-\omega^2-2i\omega+10 \right)}$$
The inverse Fourier transform leads to :
$$y(t)=\frac{1}{3}e^{-t}\sin(3t)u(t)$$
Zero state output : $\quad y(0)=0$
