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de2de:chapter-8 [2012/06/25 11:41] (current)
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 +====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:​
 +\[
 +\begin{align}
 + ​P_0(t) &= 1 \\
 + ​P_1(t) &= t \\
 + ​P_2(t) &= -\frac{1}{2} + \frac{3}{2}t^2 \\
 + ​P_3(t) &= -\frac{3}{2}t + \frac{5}{2}t^3 \\
 + ​P_4(t) &= \frac{3}{8} - \frac{15}{4}t^2 + \frac{35}{8}t^4 \\
 + ​P_5(t) &= \frac{15}{8}t - \frac{35}{4}t^3 +\frac{63}{8}t^5 \\
 +\end{align}
 +\]
 +
 +----