Solution:
See entry #3a of Table #3-2.

Solution:
See entry #14 of Table #3-2 and entry #4 of Table #3-1, which are:

14	$\displaystyle e^{-at}\cos(\omega_0 t)\;u(t)$		$\displaystyle\frac{{\bf s}+a}{({\bf s}+a)^2+\omega_0^2}$
4. Time Shift	$\displaystyle x(t-T)\;u(t-T)$		$\displaystyle e^{-T{\bf s}}\;{\bf X}({\bf s})$

Solution:
See Section 3-2.

Solution:
$\displaystyle \frac{{\bf s}^2+2{\bf s}+2}{{\bf s}^2+2{\bf s}+1}=1+\frac{1}{({\bf s}+1)^2}. $
Now use entries #1 and #6 of Table 3-2, which are:

1	$\displaystyle \delta(t)$		$\displaystyle 1$
6	$\displaystyle te^{-at}\;u(t)$		$\displaystyle \frac{1}{({\bf s}+a)^2}$

Solution:
$1+j=\sqrt{2}e^{j45^\circ }$ in entry #3 of Table 3-3, which is

3. Distinct complex

$\displaystyle \left[\frac{Ae^{j\theta}}{{\bf s}+a+jb} +\frac{Ae^{-j\theta}}{{\bf s}+a-jb}\right]$

$\displaystyle 2Ae^{-at}\cos(bt-\theta)\;u(t)$

gives $2\sqrt{2}e^{-3t}\cos(4t-45^\circ)\;u(t)$.

Solution:
$\displaystyle \frac{3}{({\bf s}+1)({\bf s}+4)}=\frac{A_1}{{\bf s}+1}+\frac{A_2}{{\bf s}+4}. $
Computation of residues $A_1$ and $A_2$:
$\displaystyle A_1=\left.({\bf s}+1)\frac3{({\bf s}+1)({\bf s}+4)}\right|_{{\bf s}=-1}=1 $
and
$\displaystyle A_2=\left.({\bf s}+4)\frac3{({\bf s}+1)({\bf s}+4)}\right|_{{\bf s}=-4}=-1. $
So $\displaystyle \frac3{({\bf s}+1)({\bf s}+4)}=\frac1{s+1}-\frac1{s+4} $
and the inverse Laplace transform is $e^{-t}\;u(t)-e^{-4t}\;u(t)$.

Solution:
Taking the Laplace transform and using entries #6 and #7 of Table 3-1 gives $\displaystyle \left[ {\bf s}^2\;{\bf Y} ({\bf s})-{\bf s} y(0)-\frac{dy}{dt}(0) \right] +2[{\bf s}\;{\bf Y} ({\bf s})-y(0)]+{\bf Y} ({\bf s})=0. $ Solving for ${\bf Y} ({\bf s})$ gives: $\displaystyle {\bf Y} ({\bf s})=\frac1{{\bf s}^2+2{\bf s}+1}=\frac1{({\bf s}+1)^2}. $
Using entry #6 of Table 3-2 gives $y(t)=te^{-t}\;u(t)$.

Solution:
$\displaystyle {\bf H} ({\bf s}) = {\cal L}\{h(t)\}=1+\frac1{{\bf s}+2}= \frac{{\bf s}+3}{{\bf s}+2}, $
$\displaystyle {\bf G} ({\bf s}) = \frac1{{\bf H} ({\bf s})}=\frac{{\bf s}+2}{{\bf s}+3}= 1-\frac1{{\bf s}+3}, $
  $\displaystyle g(t) = {\cal L}^{-1}\{{\bf G} ({\bf s})\}=\delta(t)-e^{-3t}\;u(t). $

Solution:
$\displaystyle {\bf H}({\bf s})=\frac{3{\bf s}+12}{{\bf s}^2+3{\bf s}}= \frac{{\bf Y} ({\bf s})}{{\bf X}({\bf s})}. $
Cross-multiplying gives $\displaystyle {\bf Y}({\bf s})({\bf s}^2+3{\bf s})={\bf X}({\bf s})(3{\bf s}+12) $.
Then taking the inverse Laplace transform gives $\displaystyle \frac{d^2y}{dt^2}+3\frac{dy}{dt}=3\frac{dx}{dt}+12x. $

Solution:
See the box above Example 3-13.

Solution:
$\displaystyle {\bf H}({\bf s})=C\frac{({\bf s}+3)({\bf s}+4)}{({\bf s}+1)({\bf s}+2)} =C\frac{{\bf s}^2+7{\bf s}+12}{{\bf s}^+3{\bf s}+2} $ for some $C$.
$18={\bf H} (0)=C\frac{12}{2}$, so $C=3$.
Cross-multiplying
$\displaystyle {\bf H} ({\bf s})=3\frac{{\bf s}^2+7{\bf s}+12}{{\bf s}^2+3{\bf s}+2} =\frac{{\bf Y}({\bf s})}{{\bf X}({\bf s})} $
gives
$\displaystyle {\bf Y}({\bf s})({\bf s}^2+3{\bf s}+2)= {\bf X}({\bf s})3({\bf s}^2+7{\bf s}+12). $
Taking the inverse Laplace transform gives
$\displaystyle \frac{d^2y}{dt^2}+3\frac{dy}{dt}+2y=3\frac{d^2x}{dt^2}+21\frac{dx}{dt}+36x. $

Solution:
The zero-state response is the step response $(1-e^{-t})\;u(t)$. The zero-input response is the capacitor discharging, namely $e^{-t}\;u(t)$. The total response is the sum given by $u(t)$.

Solution:
The total response can be partitioned into the sum of the zero-state response and the zero-input response. The zero-input response is simply the initial capacitor voltage decaying, which is $3e^{-t}\;u(t)$.

${\bf H}({\bf s})=\frac{{\bf s}}{{\bf s}+2}$

Solution:
From Eq. (3.122), the frequency response function is ${\bf H}({\bf s})|_{{\bf s}=j\omega}$. Here, $\displaystyle {\bf H}(j2)=\frac{j2}{2+2j}= \frac{2e^{j90^\circ }}{2\sqrt{2}e^{j45^\circ}}= \frac{e^{j45^\circ }}{\sqrt{2}}. $
The response to $\sqrt{2}\cos(2t+15^\circ )$ is then $\frac{\sqrt{2}}{\sqrt{2}}\cos(2t+15^\circ +45^\circ )=\cos(2t+60^\circ)$.

Solution:
Using properties of cascade and parallel systems interconnections: $\displaystyle h(t) = (e^{-3t}\;u(t))*(e^{-4t}\;u(t)) +\delta(t)*(e^{-4t}\;u(t)) $
        $\displaystyle = e^{-3t}\;u(t)-e^{-4t}\;u(t)+e^{-4t}\;u(t)=e^{-3t}\;u(t). $
Much easier:
$\displaystyle {\bf H}({\bf s})={\cal L}\{h(t)\}= \frac1{{\bf s}+3}\frac1{{\bf s}+4}+(1)\frac1{{\bf s}+4}= $
$\displaystyle \frac1{{\bf s}+4} \left[1+\frac1{{\bf s}+3}\right]= \frac1{{\bf s}+4}\frac{{\bf s}+4}{{\bf s}+3}=\frac1{{\bf s}+3}. $

Solution:
$\displaystyle \frac{{\bf s}+4}{({\bf s}+3)^2}=\frac1{{\bf s}+3}+\frac1{({\bf s}+3)^2}. $
So $h(t)=e^{-3t}\;u(t)+te^{-3t}\;u(t)$.

$\displaystyle {\bf H}({\bf s})=\frac{e^{-3{\bf s}}}{\bf s}$

Solution:
Integrate, then delay by 3. See entries #4 and #8 in Table 3-1, which are

4. Time shift	$\displaystyle x(t-T)\;u(t-T)$		$\displaystyle e^{-T{\bf s}}\;{\bf X}({\bf s})$
8. Time integral	$\displaystyle \int_{0^-}^t x(t')\;dt'$		$\displaystyle \frac{1}{\bf s}\;{\bf X}({\bf s})$

$h(t)$

Solution:
$\displaystyle {\bf X}({\bf s}) = {\cal L}\{e^{-3t}\;u(t)\}=\frac1{{\bf s}+3}, $
$\displaystyle {\bf Y}({\bf s}) = {\cal L}\{e^{-4t}\;u(t)\}=\frac1{{\bf s}+4}, $
$\displaystyle {\bf H}({\bf s}) = \frac{{\bf Y}({\bf s})}{{\bf X}({\bf s})}= \frac{{\bf s}+3}{{\bf s}+4}=1-\frac1{{\bf s}+4}, $
$\displaystyle h(t) = {\cal L}^{-1} \left\{ 1-\frac1{{\bf s}+4} \right\} =\delta(t)-e^{-4t}\;u(t). $

$h(t)$

Solution:
$\displaystyle {\bf X} ({\bf s}) = {\cal L}\{e^{-3t}\;u(t)\}=\frac1{{\bf s}+3}, $
$\displaystyle {\bf Y} ({\bf s}) = {\cal L}\{e^{-4t}\;u(t)\}=\frac1{{\bf s}+4}, $
$\displaystyle {\bf H} ({\bf s}) = \frac{{\bf Y} ({\bf s})}{{\bf X} ({\bf s})}=\frac{{\bf s}+3}{{\bf s}+4}=1-\frac1{{\bf s}+4}, $
$\displaystyle h(t) = {\cal L}^{-1}\{1-\frac1{{\bf s}+4}\}=\delta(t)-e^{-4t}\;u(t). $
The response to input $e^{-4t}\;u(t)$ is $\displaystyle h(t)*e^{-4t}\;u(t) =\delta(t)*e^{-4t}\;u(t)-e^{-4t}\;u(t)*e^{-4t}\;u(t) $
$\displaystyle =e^{-4t}\;u(t)-te^{-4t}\;u(t). $

Solution:
By the initial value theorem Eq. (3.31), $\displaystyle x(0+)=\lim_{{\bf s}\to\infty}{\bf s}\;{\bf X} ({\bf s}), $ which leads to $\frac1{4}$.