Solution:
The system is not time-invariant due to the time-varying coefficient $(3t)$.

Solution:
This is a linear constant-coefficient differential equation (LCCDE), so the system is LTI.

Solution:
All of these properties.

Solution:
From Eq. (2.17), the impulse response is
$\displaystyle h(t)=\frac{1}{RC}e^{-t/(RC)}\;u(t)=\frac{1}{2}e^{-t/2}\;u(t). $
The short pulse acts as an impulse with area $(10^6)1 \mu{\rm s}=1$ V-s.

$h(t)=\delta(t)-\delta(t-1)$

Solution:
From entry #6 of Table 2-1, $x(t)*\delta(t-T)=x(t-T)$.
So $y(t)=(u(t)-u(t-1))-(u(t-1)-u(t-2))$
             $=u(t)-2u(t-1)+u(t-2)$.
The system subtracts off the input delayed by 1 second from the input.

Solution:
From Eq. (2.22), $h(t)=dy_{\rm step}/dt$. From Eq. (1.18), $dr/dt=u(t)$.

$h(t)=e^{-3t}\;u(t)$

Solution:
$y(t)=h(t)*x(t)$. Compute this directly or use entry #3 in Table 2-2.

$h(t)=e^{-3t}\;u(t)$

Solution:
$y(t)=h(t)*x(t)$. Compute this directly or use entry #4 in Table 2-2.

$h(t)=u(t-1)$

Solution:
$y(t)=h(t)*x(t)$. Compute this directly or use entries #5 and #9 in Table 2-1. Then $\displaystyle x(t)*u(t)=\int_{-\infty}^tx(\tau)\;d\tau =\int_0^te^{-\tau}\;d\tau=(1-e^{-t})\;u(t). $
Then delay this by 1.

Solution:
An LTI system with impulse response $h(t)=e^{(\alpha+j\beta)t}\;u(t)$ is BIBO stable if and only if $\alpha<0$. Here $\alpha=0$, so the system is marginally stable. See Section 2-6.4.

Solution:
From Eq. (2.106), the frequency response function is ${\bf H} (\omega)=1/(j\omega+1)$.
Here $\omega=1$ (from $\sqrt{2}\cos(t)$), so $\displaystyle {\bf H} (1)=\frac{1}{1+j1}=\frac{1}{\sqrt{2}e^{j45^\circ }}. $
The $\sqrt{2}$ factors cancel.

$\frac{dy}{dt}+y(t)=\frac{dx}{dt}-x(t)$

Solution:
From Eq. (2.114), the frequency response function is ${\bf H}(\omega)=\frac{j\omega-1}{j\omega+1}$. Here $\omega=1$ (from $x(t)$), so $\displaystyle {\bf H} (1)=\frac{j-1}{j+1}=\frac{\sqrt{2}e^{j135^\circ }}{\sqrt{2}e^{j45^\circ }}=e^{j90^\circ }. $
$\cos(t+90^\circ )=-\sin(t)$.

$\displaystyle\frac{d^2y}{dt^2}+3\frac{dy}{dt}+4y(t)=6x(t)$

Solution:
From Eq. (2.114), the frequency response function is $\displaystyle {\bf H}(\omega)=\frac{6}{(j\omega)^2+3(j\omega)+4}. $
Here $\omega=2$ (from $x(t)$), so $\displaystyle {\bf H}(2)=\frac{6}{-4+j6+4}=\frac{1}{j}=e^{-j90^\circ}. $
$\cos(2t-90^\circ )=\sin(2t)$.

$\displaystyle\frac{d^2y}{dt^2}+2\frac{dy}{dt}+9y(t)=2\frac{dx}{dt}$

Solution:
From Eq. (2.114), the frequency response function is $\displaystyle {\bf H}(\omega)=\frac{2(j\omega)}{(j\omega)^2+2(j\omega)+9}. $
Since the numerator is pure imaginary, the denominator must also be pure imaginary to make ${\bf H} (\omega)$ real-valued.
This requires that $\omega=\pm3$, and ${\bf H} (\pm3)=1$, so $y(t)=x(t)$.

Solution:
See Fig. 2-29:

$\displaystyle\frac{d^2y}{dt^2}+5y=\frac{dx}{dt}+2x$

Solution:
${\bf H}(\omega) = \frac{2+j\omega}{5-\omega^2},$
${\bf H}(2) = \frac{2+j2}{5-2^2}=2+j2=2\sqrt{2}e^{j45^\circ },$
$y(t) = 2\sqrt{2}\cos(2t+45^\circ ).$

Solution:
$\displaystyle {\bf H} (\omega) = \frac{1}{j\omega+3}, $
$\displaystyle {\bf H}(3) = \frac{1}{3+j3}=\frac{1}{3\sqrt{2}}e^{-j45^\circ },$
so
$\displaystyle \frac{3\sqrt{2}}{3\sqrt{2}}\cos(3t-45^\circ )=\cos(3t-45^\circ). $

$h(t)=e^{-t}\;u(t)$

Solution:
$\displaystyle {\bf H} (\omega) = \frac{1}{j\omega+1}, $
$\displaystyle {\bf H}(0) = \frac{1}{0+1}=1. $
A constant is a 0-Hz sinusoid.

Solution:
From Eq. (2.123), the characteristic equation is ${\bf s}^2+c{\bf s}+4=0$.
The roots $\displaystyle \frac{-c\pm\sqrt{c^2-4(1)(4)}}2 $ are equal if $c=\pm 4$.
But $c=-4$ makes the system unstable.
Hence the only viable solution is $c=4$.

Solution:
If the LCCDE is overdamped, the response has form A. If underdamped, it has form B. If critically damped, it has form C. So any one of these is possible.