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 — de2de:chapter-8 [2012/06/25 11:41] (current)dcs created 2012/06/25 11:41 dcs created 2012/06/25 11:41 dcs created Line 1: Line 1: + ====Differential Equations to Difference Equations==== + ===Answers for selected problems=== + + **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) {{  fig-8-1-4-a.png ​ |Section 8.1 - Problem 4. (a)}} + + %%(c)%% {{  fig-8-1-4-c.png ​ |Section 8.1 - Problem 4. (c)}} + + (e) {{  fig-8-1-4-e.png ​ |Section 8.1 - Problem 4. (e)}} + + (g) {{  fig-8-1-4-g.png ​ |Section 8.1 - Problem 4. (g)}} + + 5. (a) {{  fig-8-1-5-a.png ​ |Section 8.1 - Problem 5. (a)}} + + %%(c)%% {{  fig-8-1-5-c.png ​ |Section 8.1 - Problem 5. (c)}} + + (e) {{  fig-8-1-5-e.png ​ |Section 8.1 - Problem 5. (e)}} + + (g) {{  fig-8-1-5-g.png ​ |Section 8.1 - Problem 5. (g)}} + + + 7. (b) {{  fig-8-1-7-b.png ​ |Section 8.1 - Problem 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) {{  fig-8-1-9-b.png ​ |Section 8.1 - Problem 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 $x(t)$: {{ fig-8-6-4-a-1.png ​ |Section 8.6 - Problem 4. (a)}} + + Phase curve: {{ fig-8-6-4-a-2.png ​ |Section 8.6 - Problem 4. (a)}} + + %%(c)%% Equations: + + \begin{align} + \dot{x} &= y \\ + \dot{y} &= -x^3 + \end{align} + + Graph of $x(t)$: {{ fig-8-6-4-c-1.png ​ |Section 8.6 - Problem 4. (c)}} + + Phase curve: {{ fig-8-6-4-c-2.png ​ |Section 8.6 - Problem 4. (c)}} + + (e) Equations: + + \begin{align} + \dot{x} &= y \\ + \dot{y} &= (x^2 - 1)y - x + \end{align} + + Graph of $x(t)$: {{ fig-8-6-4-e-1.png ​ |Section 8.6 - Problem 4. (e)}} + + Phase curve: {{ fig-8-6-4-e-2.png ​ |Section 8.6 - Problem 4. (e)}} + + 5. (a) Graph of $x(t)$ (blue) and $y(t)$ (red): {{ fig-8-6-5-a-1.png ​ |Section 8.6 - Problem 5. (a)}} + + Phase curve: {{ fig-8-6-5-a-2.png ​ |Section 8.6 - Problem 5. (a)}} + + + + %%(c)%% ​ Graph of $x(t)$ (blue) and $y(t)$ (red): {{ fig-8-6-5-c-1.png ​ |Section 8.6 - Problem 5. (c)}} + + Phase curve: {{ fig-8-6-5-c-2.png ​ |Section 8.6 - Problem 5. (c)}} + + + + + 7. Graph of $x(t)$: {{ fig-8-6-7.png ​ |Section 8.6 - Problem 7}} + + 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} + + + ----