### Differential Equations to Difference Equations

Chapter 8

Section 8.1

1. (a) $\displaystyle{x(t) = \frac{1}{3}t^3 - 2t + 3}$

(c) $\displaystyle{y(t) = \frac{2}{3}t^{\frac{3}{2}} - \frac{11}{3}}$

3. (a) $\displaystyle{x(t) = -16t^2 + 20t + 100}$

(b) $\displaystyle{t = \frac{5}{8}}$ seconds

(c) $106.25$ feet

(d) $\displaystyle{t = \frac{5}{8}(1 + \sqrt{17}) = 3.2019}$ seconds, rounded to 4 decimal places

4. (a)

(c)

(e)

(g)

5. (a)

(c)

(e)

(g)

7. (b)

(c) Yes, the object appears to be approaching a terminal speed of $1024$ feet per second.

(d) The solution, $\displaystyle{s = \frac{1024}{c^2}}$, is the terminal speed of the object since this is the speed at which the acceleration is $0$.

9. (b)

(c) $70.5$ million

(d) During the year 2022

Section 8.2

1. (a) $\displaystyle{x = 75e^{-0.9t}}$

(c) $\displaystyle{y^2 = t^2 + 25}$, or $\displaystyle{y = \sqrt{t^2 + 25}}$

(e) $\displaystyle{x^2 = 2t - 2\log(1 + t) + 16}$, or $\displaystyle{x = \sqrt{2t - 2\log(1 + t) + 16}}$

(g) $\displaystyle{x = \frac{1}{1 + 4e^{-t}}}$

2. (a) $\displaystyle{x = \frac{2x_0}{2 + x_0t^2}}$

(c) If $x_0 > 0$, domain $= (-\infty, \infty)$. If $x_0 < 0$, domain $\displaystyle{= \left\{t : t \ne \sqrt{-\frac{2}{x_0}}\right\}}$.

3. (a) The equation of the curve is $ax^2 + by^2 = c$, where $c$ is a constant. This is the equation of an ellipse, which becomes a circle if $a = b$.

4. (a) $\displaystyle{x = \frac{x_0}{1 - kx_0t}}$

(c) $\displaystyle{x = \left(\frac{1}{2}kt + \sqrt{x_0}\right)^2 = \frac{1}{4}k^2t^2 + k\sqrt{x_0}t + x_0}$

(e) The slowest population growth occurs when $0 < b < 1$, the most rapid when $b > 1$. The case $b > 1$ is called the doomsday model because the population goes to infinity in a finite amount of time.

5. (b) $\displaystyle{v = \sqrt{\frac{mg}{k}}\left(\frac{1 - e^{2\sqrt{\frac{kg}{m}}t}}{1+e^{2\sqrt{\frac{kg}{m}}t}}\right) = -\sqrt{\frac{mg}{k}}\tanh\left(\sqrt{\frac{kg}{m}}t\right)}$

Section 8.3

1. (a) $\displaystyle{x(t) = \frac{20}{9}e^{3t} - \frac{2}{3}t - \frac{2}{9}}$

(c) $\displaystyle{y = \frac{25}{2}e^{0.4t} - \frac{15}{2}}$

(e) $\displaystyle{x = t^3 + 3t^2}$

2. (a) $\displaystyle{x = 1.2e^{0.015t + 0.34875(1 - e^{-0.04t})}}$

3. (b) $\displaystyle{y = 156e^{0.02t} - 2t - 100}$

(c) $70.5$ million

(d) During the year 2022

5. (b) $\displaystyle{x = kV + V(k_0 - k)e^{-\frac{q}{V}t}}$, $\displaystyle{\lim_{t \to \infty}x(t) = kV}$

6. (b) $\displaystyle{x = \frac{2t}{1 + t^2}}$

(c) $\displaystyle{x = \frac{1}{1 + e^{-t}}}$

Section 8.4

1. (a) $\displaystyle{x = \frac{2}{3}e^t - \frac{2}{3}e^{-2t}}$

(c) $\displaystyle{x = 2e^{-t} - e^{-2t}}$

(e) $\displaystyle{x = e^t(10\cos(t) - 6\sin(t))}$

(g) $\displaystyle{x = \frac{3}{4}\sin(4t)e^{-2t}}$

(i) $\displaystyle{x = -(6 + 14t)e^{-3t}}$

3. (a) $x(0) = 0$: $\displaystyle{x = \frac{1}{3}\sin(3t)e^{-t}}$

$x(0) = -10$: $\displaystyle{x = -e^{-t}(10\cos(3t) + 3\sin(3t))}$

$x(0) = 10$: $\displaystyle{x = e^{-t}\left(10\cos(3t) + \frac{11}{3}\sin(3t)\right)}$

(b) $\dot{x}(0) = 0$: $\displaystyle{x = e^{-t}\left(10\cos(3t) + \frac{10}{3}\sin(3t)\right)}$

$\dot{x}(0) = -5$: $\displaystyle{x = e^{-t}\left(10\cos(3t) + \frac{5}{3}\sin(3t)\right)}$

$\dot{x}(0) = 5$: $\displaystyle{x = e^{-t}(10\cos(3t) + 5\sin(3t))}$

5. (a) $\displaystyle{x = c_1e^t + c_2e^{-t} + c_3e^{-2t}}$

(b) $\displaystyle{x = c_1e^{-t} + c_2te^{-t} + c_3t^2e^{-t}}$

Section 8.5

1. $g \approx 978.9$ centimeters per second per second

3. For $c = 0$, $x = 10\cos(t)$ and the motion is undamped. For $c = 10$, $x = \frac{20}{\sqrt{3}}e^{\frac{-t}{2}}\cos\left(\frac{\sqrt{3}}{2}t - \frac{\pi}{2}\right)$ and the motion is underdamped. For $c = 25$, $x = -\frac{10}{3}e^{-2t} + \frac{40}{3}e^{-\frac{t}{2}}$ and the motion is overdamped.

6. (a) Hint: Show that $x$ must satisfy the equation $\displaystyle{\ddot{x} + \frac{k}{m}x = 0}$, where $\displaystyle{k = \frac{mg}{R}}$.

(b) $84.5$ minutes

(c) $-17,614$ miles per hour

Section 8.6

1. (a) $x = c_1\cos(t) + c_2\cos(t)$, $(0, 0)$ is a center

(b) $x = c_1e^{-t} + c_2e^{-2t}$, $(0, 0)$ is a stable equilibrium

(c) $x = c_1e^t + c_2e^{-t}$, $(0, 0)$ is an unstable equilibirum

(d) $x = e^{-t}(c_1\cos(t) + c_2\sin(t))$, $(0, 0)$ is a stable equilibrium

(e) $x = e^t(c_1\cos(t) + c_2\sin(t))$, $(0, 0)$ is an unstable equilibrium

4. (a) Equations: \begin{align} \dot{x} &= y \\ \dot{y} &= -x^2 \end{align} Graph of $x(t)$:

Phase curve:

(c) Equations: \begin{align} \dot{x} &= y \\ \dot{y} &= -x^3 \end{align} Graph of $x(t)$:

Phase curve:

(e) Equations: \begin{align} \dot{x} &= y \\ \dot{y} &= (x^2 - 1)y - x \end{align} Graph of $x(t)$:

Phase curve:

5. (a) Graph of $x(t)$ (blue) and $y(t)$ (red):

Phase curve:

(c) Graph of $x(t)$ (blue) and $y(t)$ (red):

Phase curve:

7. Graph of $x(t)$:

The period is approximately $3.00$ seconds. The period of the linearized system is $2.84$ seconds. The exact period (see Problem 2 in Section 8.5) is $3.03$ seconds.

9. (a) Equations: \begin{align} \dot{x} &= y \\ \dot{y} &= -\alpha y + x(1 - x^2) \end{align}

Stationary points: $(-1, 0)$, $(0, 0)$, and $(1, 0)$

(b) $(-1, 0)$ and $(1, 0)$ are stable equilibrium points and $(0, 0)$ is an unstable equilibrium point.

(c) $(-1, 0)$, $(0, 0)$, and $(1, 0)$ are all unstable equilibrium points.

Section 8.7

1. (a) $\displaystyle{x = a_0\sum_{n=0}^\infty \frac{3^nt^n}{n!} = a_0e^{3t}}$

(c) $\displaystyle{x = a_0 + (a_0 - 1)\sum_{n=1}^\infty \frac{t^n}{n!} = 1 + (a_0 - 1)e^t}$

2. (a) For $t$ in $(-\infty, \infty)$, $x = a_0\left(1 - \frac{1}{6}t^3 + \frac{1}{180}t^6 - \cdots\right) + a_1\left(t - \frac{1}{12}t^4 + \frac{1}{504}t^7 - \cdots\right).$

(c) For $t$ in $(-\infty, \infty)$, $x = a_0\left(1 - \frac{1}{2}t^2 + \frac{1}{8}t^4 - \cdots\right) + a_1\left(t - \frac{1}{3}t^3 + \frac{1}{15}t^5 - \cdots\right).$

(e) For $t$ in $(-1, 1)$, $x = a_0\left(1 + \frac{1}{2}t^2 + \frac{7}{24}t^4 + \cdots\right) + a_1\left(t + \frac{1}{2}t^3 + \frac{13}{40}t^5 + \cdots\right).$

5.

$\mathbf{r}$ Polynomial solution
$0$ $x_1(t) = 1$
$1$ $x_2(t) = t$
$2$ $x_1(t) = 1 - 2t^2$
$3$ $x_2(t) = t - \frac{2}{3}t^3$
$4$ $x_1(t) = 1 - 4t^2 + \frac{4}{3}t^4$
$5$ $x_2(t) = t - \frac{4}{3}t^3 + \frac{4}{15}t^5$

7. (c) Legendre polynomials: \begin{align} P_0(t) &= 1 \\ P_1(t) &= t \\ P_2(t) &= -\frac{1}{2} + \frac{3}{2}t^2 \\ P_3(t) &= -\frac{3}{2}t + \frac{5}{2}t^3 \\ P_4(t) &= \frac{3}{8} - \frac{15}{4}t^2 + \frac{35}{8}t^4 \\ P_5(t) &= \frac{15}{8}t - \frac{35}{4}t^3 +\frac{63}{8}t^5 \\ \end{align}