====Differential Equations to Difference Equations====

===Answers for selected problems===

**Chapter 8**

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**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

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**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)}$

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**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}}}$

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**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}}$ 

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**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

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**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.

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**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:
\[
\begin{align}
 P_0(t) &= 1 \\
 P_1(t) &= t \\
 P_2(t) &= -\frac{1}{2} + \frac{3}{2}t^2 \\
 P_3(t) &= -\frac{3}{2}t + \frac{5}{2}t^3 \\
 P_4(t) &= \frac{3}{8} - \frac{15}{4}t^2 + \frac{35}{8}t^4 \\
 P_5(t) &= \frac{15}{8}t - \frac{35}{4}t^3 +\frac{63}{8}t^5 \\
\end{align}
\]

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