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Math 308 Week in Review 6 — Solutions

Differential Equations · Spring 2026 · Sections 3.7, 3.8, 6.1 & 6.2

Directions: Full worked solutions for all 17 questions of Week in Review 6, covering Sections 3.7, 3.8, 6.1 and 6.2. Click Solution under any problem to reveal the worked answer. Use the Contents panel or the Prev / Up / Next buttons to move between sections, and Search to find terms on this page.

Laplace Transform Reference Table

Common Laplace transform pairs
$f(t) = \mathcal{L}^{-1}\{F(s)\}$$F(s) = \mathcal{L}\{f(t)\}$
$1$$\dfrac{1}{s},\;\; s > 0$
$e^{at}$$\dfrac{1}{s-a},\;\; s > a$
$t^n \;\;(n = 1, 2, \ldots)$$\dfrac{n!}{s^{n+1}},\;\; s > 0$
$t^n e^{at}$$\dfrac{n!}{(s-a)^{n+1}},\;\; s > a$
$\sin(\omega t)$$\dfrac{\omega}{s^2+\omega^2},\;\; s > 0$
$\cos(\omega t)$$\dfrac{s}{s^2+\omega^2},\;\; s > 0$
$e^{at}\sin(\omega t)$$\dfrac{\omega}{(s-a)^2+\omega^2},\;\; s > a$
$e^{at}\cos(\omega t)$$\dfrac{s-a}{(s-a)^2+\omega^2},\;\; s > a$
Key Laplace transform properties
PropertyFormula
Linearity$\mathcal{L}\{af + bg\} = aF(s) + bG(s)$
First translation ($s$-shifting)$\mathcal{L}\{e^{at}f(t)\} = F(s-a)$
First derivative$\mathcal{L}\{f'(t)\} = sF(s) - f(0)$
Second derivative$\mathcal{L}\{f''(t)\} = s^2F(s) - sf(0) - f'(0)$

Section 3.7: Mechanical & Electrical Vibrations

1.

A mass of $m$ kg is attached to a spring with spring constant $k$ N/m and a dashpot providing a damping coefficient of $\gamma$ N·s/m. The equation of motion for free vibrations is
\[ mu'' + \gamma u' + ku = 0. \]
For each set of values, determine whether the system is overdamped, critically damped, or underdamped.
  • (a) $m = 2,\;\; \gamma = 8,\;\; k = 3$
  • (b) $m = 1,\;\; \gamma = 6,\;\; k = 9$
  • (c) $m = 2,\;\; \gamma = 4,\;\; k = 10$
  • (d) $m = 1,\;\; \gamma = 10,\;\; k = 16$
Recall. The discriminant is $\gamma^2 - 4km$. Overdamped: $\gamma^2 - 4km > 0$; critically damped: $\gamma^2 - 4km = 0$; underdamped: $\gamma^2 - 4km < 0$.
Compute the discriminant $\Delta = \gamma^2 - 4km$ for each case.
(a) $m = 2,\; \gamma = 8,\; k = 3$: $\;\Delta = 64 - 4(3)(2) = 64 - 24 = 40 > 0$. Overdamped.
(b) $m = 1,\; \gamma = 6,\; k = 9$: $\;\Delta = 36 - 4(9)(1) = 36 - 36 = 0$. Critically damped.
(c) $m = 2,\; \gamma = 4,\; k = 10$: $\;\Delta = 16 - 4(10)(2) = 16 - 80 = -64 < 0$. Underdamped.
(d) $m = 1,\; \gamma = 10,\; k = 16$: $\;\Delta = 100 - 4(16)(1) = 100 - 64 = 36 > 0$. Overdamped.
\[\boxed{(a)\;\text{Overdamped};\;\; (b)\;\text{Critically damped};\;\; (c)\;\text{Underdamped};\;\; (d)\;\text{Overdamped}}\]

2.

A mass of $1$ kg is attached to a spring with spring constant $k = 4$ N/m. There is no damping ($\gamma = 0$). The mass is displaced $3$ m in the positive direction from equilibrium and given an initial velocity of $4$ m/s in the positive direction.
  1. Write the initial value problem for $u(t)$.
  2. Solve the IVP.
  3. Express the solution in the form $u(t) = R\cos(\omega_0 t - \delta)$. Find $R$, $\omega_0$, and $\delta$.
  4. What are the natural frequency, period, and amplitude of the motion?
(a) With $m = 1$, $k = 4$, $\gamma = 0$: $\;u'' + 4u = 0$, $\;u(0) = 3$, $\;u'(0) = 4$.
(b) Characteristic equation: $r^2 + 4 = 0$, so $r = \pm 2i$. General solution:
\[u(t) = c_1\cos(2t) + c_2\sin(2t)\]
$u(0) = c_1 = 3$. $\;u'(t) = -2c_1\sin(2t) + 2c_2\cos(2t)$, so $u'(0) = 2c_2 = 4 \Rightarrow c_2 = 2$.
\[u(t) = 3\cos(2t) + 2\sin(2t)\]
(c) Write in the form $u(t) = R\cos(\omega_0 t - \delta)$, where $R = \sqrt{c_1^2 + c_2^2}$ and $\tan\delta = c_2/c_1$.
$R = \sqrt{9 + 4} = \sqrt{13}$. $\;\omega_0 = 2$.
$\tan\delta = \dfrac{2}{3}$. Since $c_1 > 0$ and $c_2 > 0$, $\delta$ is in the first quadrant: $\delta = \arctan\!\left(\dfrac{2}{3}\right)$.
(d) Natural frequency: $\omega_0 = 2$ rad/s. Period: $T = \dfrac{2\pi}{\omega_0} = \pi$. Amplitude: $R = \sqrt{13}$.
\[\boxed{u(t) = \sqrt{13}\,\cos\!\left(2t - \arctan\tfrac{2}{3}\right),\quad \omega_0 = 2,\;\; T = \pi,\;\; R = \sqrt{13}}\]

3.

A $1$-kg mass is attached to a spring with spring constant $k = 5$ N/m and a dashpot with damping coefficient $\gamma = 2$ N·s/m. The mass is released from rest at a position $4$ m from equilibrium. That is,
\[ u'' + 2u' + 5u = 0, \quad u(0) = 4,\;\; u'(0) = 0. \]
  1. Verify that the system is underdamped.
  2. Solve the initial value problem.
  3. Find the quasi-frequency $\mu$ and quasi-period $T_d$.
  4. Describe the behavior of $u(t)$ as $t \to \infty$.
(a) Characteristic equation: $r^2 + 2r + 5 = 0$. Discriminant $= 4 - 20 = -16 < 0$. Underdamped.
Roots: $r = \dfrac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i$.
(b) General solution: $u(t) = e^{-t}(c_1\cos 2t + c_2\sin 2t)$.
$u(0) = c_1 = 4$.
$u'(t) = -e^{-t}(c_1\cos 2t + c_2\sin 2t) + e^{-t}(-2c_1\sin 2t + 2c_2\cos 2t)$.
$u'(0) = -c_1 + 2c_2 = -4 + 2c_2 = 0 \;\Rightarrow\; c_2 = 2$.
\[u(t) = e^{-t}(4\cos 2t + 2\sin 2t)\]
(c) Quasi-frequency: $\mu = 2$ rad/s. Quasi-period: $T_d = \dfrac{2\pi}{\mu} = \pi$.
(d) As $t \to \infty$, $e^{-t} \to 0$, so $u(t) \to 0$. The oscillations decay exponentially to zero.
\[\boxed{u(t) = e^{-t}(4\cos 2t + 2\sin 2t),\quad \mu = 2,\quad T_d = \pi}\]

4.

Consider the critically damped system
\[ u'' + 6u' + 9u = 0, \quad u(0) = 2,\;\; u'(0) = 1. \]
  1. Verify that the system is critically damped.
  2. Solve the initial value problem.
  3. Find the time $t^*$ at which $u$ attains its maximum positive displacement, and find $u(t^*)$.
(a) Characteristic equation: $r^2 + 6r + 9 = (r + 3)^2 = 0$. Repeated root $r = -3$. Discriminant $= 36 - 36 = 0$. Critically damped.
(b) General solution: $u(t) = (c_1 + c_2 t)\,e^{-3t}$.
$u(0) = c_1 = 2$.
$u'(t) = c_2\,e^{-3t} - 3(c_1 + c_2 t)\,e^{-3t} = (c_2 - 3c_1 - 3c_2 t)\,e^{-3t}$.
$u'(0) = c_2 - 3c_1 = c_2 - 6 = 1 \;\Rightarrow\; c_2 = 7$.
\[u(t) = (2 + 7t)\,e^{-3t}\]
(c) Set $u'(t) = 0$:
\[u'(t) = (7 - 3(2 + 7t))e^{-3t} = (1 - 21t)e^{-3t} = 0\]
Since $e^{-3t} \neq 0$, we need $1 - 21t = 0$, giving $t^* = \dfrac{1}{21}$.
$u(t^*) = \left(2 + \dfrac{7}{21}\right)e^{-3/21} = \dfrac{7}{3}\,e^{-1/7} \approx 2.156$.
\[\boxed{u(t) = (2 + 7t)\,e^{-3t},\quad t^* = \tfrac{1}{21},\quad u(t^*) = \tfrac{7}{3}\,e^{-1/7}}\]

5.

A spring–mass–dashpot system has mass $m = 1$ kg and spring constant $k = 16$ N/m. The damping coefficient $\gamma > 0$ can be adjusted.
  1. For what value of $\gamma$ is the system critically damped?
  2. For what range of $\gamma$ is the system underdamped? Overdamped?
  3. The undamped natural frequency is $\omega_0 = \sqrt{k/m}$. If $\gamma = 2$, find the quasi-frequency $\mu$ and show that $\mu < \omega_0$.
(a) Critical damping occurs when $\gamma^2 - 4km = 0$, i.e. $\gamma^2 = 4(16)(1) = 64$, so $\gamma = 8$.
(b) Underdamped: $\gamma^2 < 64$, i.e. $0 < \gamma < 8$. Overdamped: $\gamma^2 > 64$, i.e. $\gamma > 8$.
(c) $\omega_0 = \sqrt{k/m} = \sqrt{16} = 4$ rad/s. With $\gamma = 2$:
Characteristic roots: $r = \dfrac{-2 \pm \sqrt{4 - 64}}{2} = -1 \pm i\sqrt{15}$.
Quasi-frequency: $\mu = \sqrt{15} \approx 3.873$ rad/s. Since $\sqrt{15} < 4 = \omega_0$, we confirm $\mu < \omega_0$.
(Damping always reduces the frequency of oscillation.)
\[\boxed{\gamma_{\text{crit}} = 8;\quad \text{underdamped: } 0 < \gamma < 8;\quad \text{overdamped: } \gamma > 8;\quad \mu = \sqrt{15} < 4 = \omega_0}\]

Section 3.8: Forced Mechanical Vibrations

6.

A mass weighing $10$ N stretches a spring $2$ m. The system has a damping force numerically equal to $2$ times the instantaneous velocity. An external force $F(t) = 10\cos t$ N is applied to the mass, which is initially at rest in the equilibrium position.
  1. Using $g = 10$ m/s², find $m$, $k$, and $\gamma$, and write the equation of motion.
  2. Solve the initial value problem.
  3. Identify the transient part and the steady-state part of the solution.
  4. Find the amplitude $R$ and phase $\delta$ of the steady-state solution. That is, write the steady-state as $U(t) = R\cos(\omega t - \delta)$.
Recall. Weight $W = mg$, so $m = W/g$. Hooke's law gives $k = W/L$ where $L$ is the static deflection. The steady-state solution $U(t) = A\cos\omega t + B\sin\omega t$ can be written as $R\cos(\omega t - \delta)$ where $R = \sqrt{A^2 + B^2}$ and $\tan\delta = B/A$.
(a) $m = W/g = 10/10 = 1$ kg. $\;k = W/L = 10/2 = 5$ N/m. $\;\gamma = 2$. The equation of motion is:
\[u'' + 2u' + 5u = 10\cos t, \quad u(0) = 0,\;\; u'(0) = 0\]
(b) Characteristic equation: $r^2 + 2r + 5 = 0$, so $r = -1 \pm 2i$.
Homogeneous solution: $u_c(t) = e^{-t}(c_1\cos 2t + c_2\sin 2t)$.
For the particular solution, try $u_p = A\cos t + B\sin t$:
$u_p' = -A\sin t + B\cos t$, $\;u_p'' = -A\cos t - B\sin t$.
Substituting into $u'' + 2u' + 5u = 10\cos t$:
\[(-A + 2B + 5A)\cos t + (-B - 2A + 5B)\sin t = 10\cos t\]
\[(4A + 2B)\cos t + (-2A + 4B)\sin t = 10\cos t\]
$\cos t$: $4A + 2B = 10$. $\;\sin t$: $-2A + 4B = 0 \Rightarrow A = 2B$.
Substituting: $8B + 2B = 10 \Rightarrow B = 1$, $A = 2$. So $u_p = 2\cos t + \sin t$.
General solution: $u = e^{-t}(c_1\cos 2t + c_2\sin 2t) + 2\cos t + \sin t$.
$u(0) = c_1 + 2 = 0 \Rightarrow c_1 = -2$.
$u'(0) = -c_1 + 2c_2 + 1 = 2 + 2c_2 + 1 = 0 \Rightarrow c_2 = -\dfrac{3}{2}$.
\[u(t) = e^{-t}\!\left(-2\cos 2t - \tfrac{3}{2}\sin 2t\right) + 2\cos t + \sin t\]
(c) Transient: $u_{tr}(t) = e^{-t}\!\left(-2\cos 2t - \tfrac{3}{2}\sin 2t\right) \to 0$ as $t \to \infty$.
Steady-state: $U(t) = 2\cos t + \sin t$.
(d) $R = \sqrt{A^2 + B^2} = \sqrt{4 + 1} = \sqrt{5}$.
$\tan\delta = B/A = 1/2$. Since $A > 0$, $B > 0$, $\delta$ is in the first quadrant: $\delta = \arctan(1/2)$.
\[\boxed{U(t) = \sqrt{5}\,\cos(t - \arctan(1/2)),\quad R = \sqrt{5},\;\; \delta = \arctan(1/2)}\]

7.

Consider the undamped forced vibration problem
\[ u'' + 9u = 5\cos(2t), \quad u(0) = 0,\;\; u'(0) = 0. \]
  1. Show that the natural frequency of the system is $\omega_0 = 3$ and the forcing frequency is $\omega = 2$.
  2. Solve the initial value problem.
  3. Use the identity $\cos A - \cos B = -2\sin\!\Big(\dfrac{A+B}{2}\Big)\sin\!\Big(\dfrac{A-B}{2}\Big)$ to express the solution as a product of sines and identify the beating phenomenon.
  4. What would happen if the forcing frequency equaled the natural frequency ($\omega = \omega_0 = 3$)? Without solving, describe the behavior and name the phenomenon.
Note. Beating occurs when $\omega$ is close to but not equal to $\omega_0$. The solution oscillates rapidly with a slowly varying amplitude envelope.
(a) The equation $u'' + 9u = 5\cos 2t$ has $\omega_0^2 = 9$, so $\omega_0 = 3$ (natural). The forcing frequency is $\omega = 2$. Since $\omega \neq \omega_0$, this is not resonance.
(b) Homogeneous solution: $u_c = c_1\cos 3t + c_2\sin 3t$.
Particular solution: try $u_p = A\cos 2t + B\sin 2t$.
$u_p'' + 9u_p = (-4A + 9A)\cos 2t + (-4B + 9B)\sin 2t = 5A\cos 2t + 5B\sin 2t = 5\cos 2t$.
So $A = 1$, $B = 0$. Thus $u_p = \cos 2t$.
General solution: $u = c_1\cos 3t + c_2\sin 3t + \cos 2t$.
$u(0) = c_1 + 1 = 0 \;\Rightarrow\; c_1 = -1$.
$u'(0) = 3c_2 = 0 \;\Rightarrow\; c_2 = 0$.
\[u(t) = \cos 2t - \cos 3t\]
(c) Using the identity $\cos A - \cos B = -2\sin\!\left(\dfrac{A+B}{2}\right)\sin\!\left(\dfrac{A-B}{2}\right)$ with $A = 2t$, $B = 3t$:
\[u(t) = -2\sin\!\left(\frac{5t}{2}\right)\sin\!\left(\frac{-t}{2}\right) = 2\sin\!\left(\frac{t}{2}\right)\sin\!\left(\frac{5t}{2}\right)\]
This is the product of a slowly varying envelope $2\sin(t/2)$ and a rapid oscillation $\sin(5t/2)$. This is the beating phenomenon: the amplitude oscillates periodically with beat frequency $|\omega_0 - \omega|/2 = 1/2$.
(d) If $\omega = \omega_0 = 3$, the forcing frequency matches the natural frequency. The particular solution would contain a $t\sin 3t$ term (the MoUC trial must be multiplied by $t$). This gives unbounded oscillations whose amplitude grows linearly with time. This is pure resonance.
\[\boxed{u(t) = 2\sin(t/2)\sin(5t/2) \quad\text{(beating)}}\]

Section 6.1: Definition of the Laplace Transform

8.

Use the definition $\displaystyle\mathcal{L}\{f(t)\} = \int_0^{\infty} e^{-st}\,f(t)\,dt$ to compute each Laplace transform. State for which values of $s$ the integral converges.
  • (a) $f(t) = 5$
  • (b) $f(t) = e^{-2t}$
  • (c) $f(t) = 3t$
(a) $f(t) = 5$:
\[\mathcal{L}\{5\} = \int_0^{\infty} 5e^{-st}\,dt = 5\left[-\frac{e^{-st}}{s}\right]_0^{\infty} = 5\left(0 + \frac{1}{s}\right) = \frac{5}{s}, \quad s > 0\]
(b) $f(t) = e^{-2t}$:
\[\mathcal{L}\{e^{-2t}\} = \int_0^{\infty} e^{-(s+2)t}\,dt = \left[-\frac{e^{-(s+2)t}}{s+2}\right]_0^{\infty} = \frac{1}{s+2}, \quad s > -2\]
(c) $f(t) = 3t$. Integrate by parts with $u = 3t$, $dv = e^{-st}\,dt$:
\[\mathcal{L}\{3t\} = \left[-\frac{3t\,e^{-st}}{s}\right]_0^{\infty} + \int_0^{\infty} \frac{3}{s}\,e^{-st}\,dt = 0 + \frac{3}{s}\cdot\frac{1}{s} = \frac{3}{s^2}, \quad s > 0\]
\[\boxed{(a)\;\frac{5}{s},\; s > 0;\quad (b)\;\frac{1}{s+2},\; s > -2;\quad (c)\;\frac{3}{s^2},\; s > 0}\]

9.

Use the table of Laplace transforms (Table 6.2.1) and the linearity property to find $\mathcal{L}\{f(t)\}$.
  • (a) $f(t) = 4e^{3t} - 2\sin(5t) + 7$
  • (b) $f(t) = t^4 + 3t^2 - 1$
  • (c) $f(t) = e^{-2t}\cos(3t)$
  • (d) $f(t) = 5te^{4t}$
Recall. First Translation Theorem (Theorem 6.1.1 Corollary): $\mathcal{L}\{e^{at}f(t)\} = F(s - a)$ where $F(s) = \mathcal{L}\{f(t)\}$.
(a) $f(t) = 4e^{3t} - 2\sin(5t) + 7$. By linearity:
\[\mathcal{L}\{f\} = \frac{4}{s-3} - \frac{2\cdot 5}{s^2+25} + \frac{7}{s} = \frac{4}{s-3} - \frac{10}{s^2+25} + \frac{7}{s}\]
(b) $f(t) = t^4 + 3t^2 - 1$. Using $\mathcal{L}\{t^n\} = n!/s^{n+1}$:
\[\mathcal{L}\{f\} = \frac{4!}{s^5} + \frac{3\cdot 2!}{s^3} - \frac{1}{s} = \frac{24}{s^5} + \frac{6}{s^3} - \frac{1}{s}\]
(c) $f(t) = e^{-2t}\cos(3t)$. First translation with $a = -2$: start with $\mathcal{L}\{\cos 3t\} = \dfrac{s}{s^2+9}$, then replace $s$ by $s+2$:
\[\mathcal{L}\{e^{-2t}\cos 3t\} = \frac{s+2}{(s+2)^2+9}\]
(d) $f(t) = 5te^{4t}$. Start with $\mathcal{L}\{t\} = \dfrac{1}{s^2}$, then first translation with $a = 4$: replace $s$ by $s - 4$:
\[\mathcal{L}\{5te^{4t}\} = \frac{5}{(s-4)^2}\]
\[\boxed{(a)\;\frac{4}{s-3} - \frac{10}{s^2+25} + \frac{7}{s};\quad (b)\;\frac{24}{s^5} + \frac{6}{s^3} - \frac{1}{s};\quad (c)\;\frac{s+2}{(s+2)^2+9};\quad (d)\;\frac{5}{(s-4)^2}}\]

Section 6.2: Solution of Initial Value Problems

10.

Find the inverse Laplace transform $f(t) = \mathcal{L}^{-1}\{F(s)\}$.
  • (a) $\displaystyle F(s) = \frac{3}{s} + \frac{2}{s^3}$
  • (b) $\displaystyle F(s) = \frac{4}{s + 1}$
  • (c) $\displaystyle F(s) = \frac{5}{s^2 + 25}$
  • (d) $\displaystyle F(s) = \frac{3s}{s^2 + 16}$
(a) $F(s) = \dfrac{3}{s} + \dfrac{2}{s^3}$. Using $\mathcal{L}^{-1}\{1/s\} = 1$ and $\mathcal{L}^{-1}\{n!/s^{n+1}\} = t^n$:
$\dfrac{2}{s^3} = \dfrac{2}{2!}\cdot\dfrac{2!}{s^3}$, so $\mathcal{L}^{-1}\!\left\{\dfrac{2}{s^3}\right\} = t^2$.
\[f(t) = 3 + t^2\]
(b) $F(s) = \dfrac{4}{s+1} = \dfrac{4}{s - (-1)}$, so $f(t) = 4e^{-t}$.
(c) $F(s) = \dfrac{5}{s^2+25} = \dfrac{5}{s^2+5^2}$. This matches $\dfrac{\omega}{s^2+\omega^2}$ with $\omega = 5$, so $f(t) = \sin(5t)$.
(d) $F(s) = \dfrac{3s}{s^2+16} = 3\cdot\dfrac{s}{s^2+4^2}$, so $f(t) = 3\cos(4t)$.
\[\boxed{(a)\; 3 + t^2;\quad (b)\; 4e^{-t};\quad (c)\; \sin 5t;\quad (d)\; 3\cos 4t}\]

11.

Use partial fractions to find $\mathcal{L}^{-1}\{F(s)\}$.
  • (a) $\displaystyle F(s) = \frac{2s + 10}{(s + 2)(s + 4)}$
  • (b) $\displaystyle F(s) = \frac{6}{s(s^2 + 4)}$
(a) $F(s) = \dfrac{2s+10}{(s+2)(s+4)}$. Set up partial fractions:
\[\frac{2s+10}{(s+2)(s+4)} = \frac{A}{s+2} + \frac{B}{s+4}\]
Multiply by $(s+2)(s+4)$: $\;2s + 10 = A(s+4) + B(s+2)$.
$s = -2$: $-4 + 10 = 2A \;\Rightarrow\; A = 3$.
$s = -4$: $-8 + 10 = -2B \;\Rightarrow\; B = -1$.
\[f(t) = 3e^{-2t} - e^{-4t}\]
(b) $F(s) = \dfrac{6}{s(s^2+4)}$. Set up partial fractions:
\[\frac{6}{s(s^2+4)} = \frac{A}{s} + \frac{Bs + C}{s^2+4}\]
Multiply by $s(s^2+4)$: $\;6 = A(s^2+4) + (Bs+C)s$.
$s = 0$: $6 = 4A \;\Rightarrow\; A = 3/2$.
Coeff. of $s^2$: $0 = A + B \;\Rightarrow\; B = -3/2$.
Coeff. of $s^1$: $0 = C$.
\[F(s) = \frac{3/2}{s} - \frac{3s/2}{s^2+4} = \frac{3}{2}\cdot\frac{1}{s} - \frac{3}{2}\cdot\frac{s}{s^2+4}\]
\[\boxed{(a)\; 3e^{-2t} - e^{-4t};\quad (b)\; \tfrac{3}{2} - \tfrac{3}{2}\cos 2t}\]

12.

Find $\mathcal{L}^{-1}\{F(s)\}$ by completing the square in the denominator.
  • (a) $\displaystyle F(s) = \frac{2s + 5}{s^2 + 4s + 13}$
  • (b) $\displaystyle F(s) = \frac{3}{s^2 + 2s + 10}$
Hint. Write $s^2 + 4s + 13 = (s+2)^2 + 9$ and match to the forms $\displaystyle\frac{s - a}{(s - a)^2 + b^2}$ and $\displaystyle\frac{b}{(s - a)^2 + b^2}$.
(a) $F(s) = \dfrac{2s+5}{s^2+4s+13}$. Complete the square: $s^2+4s+13 = (s+2)^2 + 9$.
Rewrite the numerator in terms of $(s+2)$: $\;2s + 5 = 2(s+2) + 1$.
\[\frac{2(s+2) + 1}{(s+2)^2+9} = 2\cdot\frac{s+2}{(s+2)^2+9} + \frac{1}{3}\cdot\frac{3}{(s+2)^2+9}\]
Inverting (with $a = -2$ and $\beta = 3$):
\[f(t) = 2e^{-2t}\cos 3t + \tfrac{1}{3}\,e^{-2t}\sin 3t\]
(b) $F(s) = \dfrac{3}{s^2+2s+10}$. Complete the square: $s^2+2s+10 = (s+1)^2 + 9$.
\[\frac{3}{(s+1)^2+9} = 1\cdot\frac{3}{(s+1)^2+9}\]
This matches $\dfrac{\beta}{(s-a)^2+\beta^2}$ with $a = -1$, $\beta = 3$:
\[\boxed{(a)\; 2e^{-2t}\cos 3t + \tfrac{1}{3}e^{-2t}\sin 3t;\quad (b)\; e^{-t}\sin 3t}\]

13.

Use the Laplace transform to solve the initial value problem.
\[ y'' + 4y = 0, \quad y(0) = 2,\;\; y'(0) = -6. \]
Step 1: Take $\mathcal{L}$ of both sides. Using $\mathcal{L}\{y''\} = s^2Y - sy(0) - y'(0)$:
\[[s^2Y - 2s + 6] + 4Y = 0\]
Step 2: Solve for $Y(s)$:
\[Y(s^2+4) = 2s - 6 \quad\Rightarrow\quad Y(s) = \frac{2s - 6}{s^2+4} = 2\cdot\frac{s}{s^2+4} - 3\cdot\frac{2}{s^2+4}\]
Step 3: Invert term by term using the table:
\[\boxed{y(t) = 2\cos 2t - 3\sin 2t}\]

14.

Use the Laplace transform to solve the initial value problem.
\[ y'' - 4y' + 3y = 0, \quad y(0) = 1,\;\; y'(0) = 5. \]
Step 1: Take $\mathcal{L}$:
\[[s^2Y - s - 5] - 4[sY - 1] + 3Y = 0\]
\[s^2Y - s - 5 - 4sY + 4 + 3Y = 0\]
Step 2: Collect terms:
\[Y(s^2 - 4s + 3) = s + 1 \quad\Rightarrow\quad Y(s) = \frac{s+1}{(s-1)(s-3)}\]
Step 3: Partial fractions. $\dfrac{s+1}{(s-1)(s-3)} = \dfrac{A}{s-1} + \dfrac{B}{s-3}$.
Multiply by $(s-1)(s-3)$: $\;s + 1 = A(s-3) + B(s-1)$.
$s = 1$: $2 = -2A \;\Rightarrow\; A = -1$. $\;s = 3$: $4 = 2B \;\Rightarrow\; B = 2$.
\[\boxed{y(t) = -e^{t} + 2e^{3t}}\]

15.

Use the Laplace transform to solve the initial value problem.
\[ y'' - 2y' - 3y = 6, \quad y(0) = 0,\;\; y'(0) = 0. \]
Recall. $\mathcal{L}\{y''\} = s^2\,Y(s) - s\,y(0) - y'(0)$ and $\mathcal{L}\{y'\} = s\,Y(s) - y(0)$.
Step 1: Take $\mathcal{L}$. With $y(0) = y'(0) = 0$: $\;s^2Y - 2sY - 3Y = \dfrac{6}{s}$.
Step 2: $Y(s^2 - 2s - 3) = \dfrac{6}{s}$, so $Y(s) = \dfrac{6}{s(s^2-2s-3)} = \dfrac{6}{s(s-3)(s+1)}$.
Step 3: Partial fractions:
\[\frac{6}{s(s-3)(s+1)} = \frac{A}{s} + \frac{B}{s-3} + \frac{C}{s+1}\]
Multiply by $s(s-3)(s+1)$: $\;6 = A(s-3)(s+1) + Bs(s+1) + Cs(s-3)$.
$s = 0$: $6 = A(-3)(1) \;\Rightarrow\; A = -2$.
$s = 3$: $6 = B(3)(4) \;\Rightarrow\; B = 1/2$.
$s = -1$: $6 = C(-1)(-4) \;\Rightarrow\; C = 3/2$.
\[\boxed{y(t) = -2 + \tfrac{1}{2}\,e^{3t} + \tfrac{3}{2}\,e^{-t}}\]

16.

Use the Laplace transform to solve the initial value problem.
\[ y'' + 2y' + 5y = 0, \quad y(0) = 2,\;\; y'(0) = -2. \]
Hint. After taking the Laplace transform, you will need to complete the square in the denominator.
Step 1: Take $\mathcal{L}$:
\[[s^2Y - 2s + 2] + 2[sY - 2] + 5Y = 0\]
\[Y(s^2 + 2s + 5) = 2s + 2\]
Step 2: $Y(s) = \dfrac{2s + 2}{s^2 + 2s + 5}$. Complete the square: $s^2 + 2s + 5 = (s+1)^2 + 4$.
Rewrite the numerator: $\;2s + 2 = 2(s + 1)$.
\[Y(s) = \frac{2(s+1)}{(s+1)^2 + 4}\]
Step 3: This matches $\dfrac{s - a}{(s-a)^2 + \beta^2}$ with $a = -1$, $\beta = 2$:
\[\boxed{y(t) = 2e^{-t}\cos 2t}\]

17.

Use the Laplace transform to solve the initial value problem.
\[ y'' + 2y' + 5y = 4e^{-t}, \quad y(0) = 1,\;\; y'(0) = 0. \]
Hint. After taking the Laplace transform, you will need partial fractions and completing the square. Note that $s^2 + 2s + 5 = (s + 1)^2 + 4$.
Step 1: Take $\mathcal{L}$. Using $y(0) = 1$, $y'(0) = 0$:
\[[s^2Y - s] + 2[sY - 1] + 5Y = \frac{4}{s+1}\]
\[Y(s^2+2s+5) = s + 2 + \frac{4}{s+1} = \frac{(s+2)(s+1) + 4}{s+1} = \frac{s^2 + 3s + 6}{s+1}\]
Step 2: $Y(s) = \dfrac{s^2+3s+6}{(s+1)(s^2+2s+5)}$. Note $s^2+2s+5 = (s+1)^2+4$.
Set up partial fractions:
\[\frac{s^2+3s+6}{(s+1)[(s+1)^2+4]} = \frac{A}{s+1} + \frac{B(s+1)+C}{(s+1)^2+4}\]
Multiply through by $(s+1)[(s+1)^2+4]$:
\[s^2+3s+6 = A[(s+1)^2+4] + [B(s+1)+C](s+1)\]
$s = -1$: $1 - 3 + 6 = A(0 + 4) = 4A \;\Rightarrow\; A = 1$.
Expand and let $w = s + 1$ (so $s^2+3s+6 = w^2 + w + 4$):
\[w^2 + w + 4 = (w^2 + 4) + Bw^2 + Cw = (1+B)w^2 + Cw + 4\]
$w^2$: $1 = 1 + B \;\Rightarrow\; B = 0$. $\;w^1$: $1 = C$.
\[Y(s) = \frac{1}{s+1} + \frac{1}{(s+1)^2+4} = \frac{1}{s+1} + \frac{1}{2}\cdot\frac{2}{(s+1)^2+4}\]
Step 3: Invert each term:
\[\boxed{y(t) = e^{-t} + \tfrac{1}{2}\,e^{-t}\sin 2t = e^{-t}\!\left(1 + \tfrac{1}{2}\sin 2t\right)}\]
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