Electrical – Finding the zero-state response

I do not believe there is any problem with the notation used by the author. He just wanted to express the following (\$x\$ input, \$y\$ output - the response to a eigenfunction \$e^{st}\$):

$$y(t) = \int_{-\infty}^\infty h(\tau)x(t-\tau)d\tau $$ $$ = \int_{-\infty}^\infty h(\tau)e^{s(t-\tau)}d\tau $$ $$ = e^{st} \int_{-\infty}^\infty h(\tau)e^{-s\tau}d\tau $$

Thus, the response to \$e^{st}\$ is of the form

$$y(t) = H(s)e^{st}$$

where \$H(s)\$ is a complex constant whose value depends on \$s\$ and wich is related to the impulse response by

$$ H(s) = \int_{-\infty}^\infty h(\tau)e^{-s\tau}d\tau $$

There is no conflict time x frequency here.

Also, note:

If \$\lambda\$ is a pole of \$H(s)\$, the mode \$e^{{\lambda}t}\$ can be generated at the output by an initial state without the application of any input. If \$\lambda\$ is not a pole of \$H(s)\$, then this is not possible. In this case, the only way to generate \$e^{{\lambda}t}\$ at the output is to apply \$e^{{\lambda}t}\$ at the input. In other hand, if the input \$x(t)\$ is of the form \$e^{{\lambda}t}\$, where \$\lambda\$ is not a pole of \$H(s)\$, then the output due to input \$x(t) = e^{{\lambda}t}\$ is equal to \$y(t) = H(\lambda)e^{{\lambda}t}\$.

There seems to be an error in the approach of the transfer function:

$$\frac{V(s)}{V_s(s)}=\frac{\frac{1}{sC}}{R_1+\frac{1}{sC}||(R_2+sL)}$$

when the correct is:

$$ \frac{V(s)}{V_s(s)}=\frac{\frac{1}{sC}||(R_2+sL)}{R_1+\frac{1}{sC}||(R_2+sL)} $$

according to the voltage divider.