Chapter 4
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Chapter 4
Calculus Volume 2Chapter 4
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Chapter 4
Checkpoint
4.2
$5$
4.3
$y = 2x^{2} + 3x + 2$
4.5
$y = \frac{1}{3}x^{3} - 2x^{2} + 3x - 6e^{x} + 14$
4.6
$v(t) = -9.8t$
4.7
4.8
The equilibrium solutions are $y = -2$ and $y = 2.$ For this equation, $y = -2$ is an unstable equilibrium solution, and $y = 2$ is a semi-stable equilibrium solution.
4.9
| $n$ | $x_{n}$ | $y_{n} = y_{n - 1} + hf(x_{n - 1},y_{n - 1})$ |
|------|---------|-----------------------------------------------|
| $0$ | $1$ | $-2$ |
| $1$ | $1.1$ | $y_{1} = y_{0} + hf(x_{0},y_{0}) = -1.5$ |
| $2$ | $1.2$ | $y_{2} = y_{1} + hf(x_{1},y_{1}) = -1.1419$ |
| $3$ | $1.3$ | $y_{3} = y_{2} + hf(x_{2},y_{2}) = -0.8387$ |
| $4$ | $1.4$ | $y_{4} = y_{3} + hf(x_{3},y_{3}) = -0.5487$ |
| $5$ | $1.5$ | $y_{5} = y_{4} + hf(x_{4},y_{4}) = -0.2442$ |
| $6$ | $1.6$ | $y_{6} = y_{5} + hf(x_{5},y_{5}) = 0.0993$ |
| $7$ | $1.7$ | $y_{7} = y_{6} + hf(x_{6},y_{6}) = 0.5099$ |
| $8$ | $1.8$ | $y_{8} = y_{7} + hf(x_{7},y_{7}) = 1.0272$ |
| $9$ | $1.9$ | $y_{9} = y_{8} + hf(x_{8},y_{8}) = 1.7159$ |
| $10$ | $2$ | $y_{10} = y_{9} + hf(x_{9},y_{9}) = 2.6962$ |
4.10
$y = 2 + Ce^{x^{2} + 3x}$
4.11
$y = \frac{4 + 14e^{x^{2} + x}}{1 - 7e^{x^{2} + x}}$
4.12
Initial value problem:
$\frac{du}{dt} = 2.4 - \frac{2u}{25},\quad u(0) = 3$
$\text{Solution:}\ u(t) = 30 - 27e^{\text{−}2t\text{/}25}$
$\text{Concentration:~}30 - 27e^{\frac{- 2\mathit{t}}{25}}$
4.13
1. Initial value problem
$\frac{dT}{dt} = k\left( {T - 70} \right),\quad T(0) = 450$
2. $T(t) = 70 + 380e^{kt}$
3. Approximately $114$ minutes.
4.14
1. $\frac{dP}{dt} = 0.04\left( {1 - \frac{P}{750}} \right),\quad P(0) = 200$
2.
3. $P(t) = \frac{3000e^{.04t}}{11 + 4e^{.04t}}$
4. After $12$ months, the population will be $P(12) \approx 278$ rabbits.
4.15
$y\prime + \frac{15}{x + 3}y = \frac{10x - 20}{x + 3};p(x) = \frac{15}{x + 3}$ and $q(x) = \frac{10x - 20}{x + 3}$
4.16
$y = \frac{x^{3} + x^{2} + C}{x - 2}$
4.17
$y = - 2x - \frac{5}{2} + \frac{1}{2}e^{2x}$
4.18
1. $\begin{array}{rll}
\frac{dv}{dt} & = & {\text{−}v - 9.8} \\
{v(0)} & = & 0
\end{array}$
2. $v(t) = 9.8\left( {e^{\text{−}t} - 1} \right)$
3. $\underset{t\rightarrow\infty}{\text{lim}}v(t) = \underset{t\rightarrow\infty}{\text{lim}}\left( {9.8\left( {e^{\text{−}t} - 1} \right)} \right) = -9.8\ \text{m/s} \approx - 21.922\ \text{mph}$
4.19
Initial-value problem:
$8q^{\prime} + \frac{1}{0.02}q = 20\mspace{2mu}\text{sin}\mspace{2mu} 5t,\quad q(0) = 4$
$q(t) = \frac{10\mspace{2mu}\text{sin}\mspace{2mu} 5t - 8\mspace{2mu}\text{cos}\mspace{2mu} 5t + 172e^{-6.25t}}{41}$
Section 4.1 Exercises
1.
$1$
3.
$3$
5.
$1$
7.
$1$
19.
$y = 4 + \frac{3x^{4}}{4}$
21.
$y = \frac{1}{2}e^{x^{2}}$
23.
$y = 2e^{\text{−}{1\text{/}x}}$
25.
$u = \text{sin}^{-1}\left( e^{-1 + t} \right)$
27.
$y = - \frac{\sqrt{x + 1}}{\sqrt{1 - x}} - 1$
29.
$y = C - x + x\mspace{2mu}\text{ln}\mspace{2mu} x - \text{ln}(\text{cos}\mspace{2mu} x)$
31.
$y = C + \frac{4^{x}}{\text{ln}(4)}$
33.
$y = \frac{2}{3}\sqrt{t^{2} + 16}\left( {t^{2} + 16} \right) + C$
35.
$x = \frac{2}{15}\sqrt{4 + t}\left( {3t^{2} + 4t - 32} \right) + C$
37.
$y = Cx$
39.
$y = 1 - \frac{t^{2}}{2},y = - \frac{t^{2}}{2} - 1$
41.
$y = e^{\text{−}t},y = \text{−}e^{\text{−}t}$
43.
$y = 2\left( {t^{2} + 5} \right),t = 3\sqrt{5}$
45.
$y = 10e^{-2t},t = - \frac{1}{2}\mspace{2mu}\text{ln}\mspace{2mu}\left( \frac{1}{10} \right)$
47.
$y = \frac{1}{4}\left( {41 - e^{-4t}} \right),$ never
49.
Solution changes from increasing to decreasing at $y(0) = 0$
51.
Solution changes from increasing to decreasing at $y(0) = 0$
53.
$v(t) = -32t + a$
55.
$0$ ft/s
57.
$52.354$ meters
59.
$x = 50t - \frac{15}{\pi^{2}}\text{cos}(\pi t) + \frac{3}{\pi^{2}},2$ hours $1$ minute
61.
$y = 4e^{3t}$
63.
$y = 3 - 2t + t^{2}$
65.
$y = \frac{1}{k}\left( {e^{kt} - 1} \right)$ and $y = x$
Section 4.2 Exercises
67.
69.
$y = 0$ is a stable equilibrium
71.
73.
$y = 0$ is a stable equilibrium and $y = 2$ is unstable
75.
General solution is $y = e^{t} + C$.
77.
General solution is $y = e^{t}(t - 1) + C$.
79.
81.
83.
85.
E
87.
A
89.
B
91.
A
93.
C
95.
$2.24,$ exact: $2$
97.
$7.739264,$ exact: $5(e - 1)$
99.
$-0.2535$ exact: $0$
101.
$1.345,$ exact: $\frac{1}{\text{ln}(2)}$
103.
$-4,$ exact: $\text{−}{1\text{/}2}$
105.
107.
$y\prime = 2e^{t^{2}\text{/}2}$
109.
$2$
111.
$3.2756$
113.
$2\sqrt{e}$
| Step Size | Relative Error |
|--------------|----------------|
| $h = 0.1$ | $0.3935$ |
| $h = 0.01$ | $0.06163$ |
| $h = 0.001$ | $0.006612$ |
| $h = 0.0001$ | $0.0006661$ |
115.
117.
$4.0741e^{-10}$
Section 4.3 Exercises
119.
$y = e^{t} - 1$
121.
$y = 1 + Ce^{\text{−}t}$
123.
$y = Cxe^{-1\text{/}x}$
125.
$y = \frac{1}{C - x^{2}}$
127.
$y = - \frac{2}{C + \text{ln}\mspace{2mu} x}$
129.
$y = Ce^{x}\left( {x + 1} \right) + 1$
131.
$y = \text{sin}\left( {\text{ln}\mspace{2mu} t + C} \right)$
133.
$y = \text{−}\text{ln}(e^{\text{−}x})$
135.
$y = \frac{1}{\sqrt{2 - e^{x^{2}}}}$
137.
$y = \text{tanh}^{-1}\left( \frac{x^{2}}{2} \right)$
139.
$x = \text{sin}\left( {1 - t + t\mspace{2mu}\text{ln}\mspace{2mu} t} \right)$
141.
$y = \text{ln}(\text{ln}(5)) - \text{ln}(2 - 5^{x})$
143.
$y = Ce^{-2x} + \frac{1}{2}$
145.
$y = \frac{1}{\sqrt{2}\sqrt{C - e^{x}}}$
147.
$y = Ce^{\text{−}x}x^{x}$
149.
$y = \frac{r}{d}\left( {1 - e^{\text{−}dt}} \right)$
151.
$y(t) = 10 - 9e^{\text{−}{x\text{/}50}}$
153.
$134.3$ kilograms
155.
$720$ seconds
157.
$24$ hours $57$ minutes
159.
$T(t) = 20 + 50e^{-0.125t}$
161.
$T(t) = 20 + 38.5e^{-0.125t}$
163.
$y = \left( {c + \frac{b}{a}} \right)e^{ax} - \frac{b}{a}$
165.
$y(t) = cL + (I - cL)e^{\text{−}{{rt}\text{/}L}}$
167.
$y = 40\left( {1 - e^{-0.1t}} \right),40$ g/cm2
Section 4.4 Exercises
169.
$P = 0$ semi-stable
171.
$P = \frac{10e^{10x}}{e^{10x} + 4}$
173.
$P(t) = \frac{10000e^{0.02t}}{150 + 50e^{0.02t}}$
175.
$69$ hours $5$ minutes
177.
$8$ years $11$ months
179.
181.
$P_{1}$ semi-stable
183.
$P_{2} > 0$ stable
185.
$P_{1} = 0$ is semi-stable
187.
$P(t) = \frac{3500}{\left( {4 + 3e^{- 035t}} \right)}$
189.
191.
$P(t) = \frac{850 + 500e^{0.009t}}{85 + 5e^{0.009t}}$
193.
$13$ years months
195.
197.
$31.465$ days
199.
September $2008$
201.
$\frac{K + T}{2}$
203.
$r = 0.0405$
205.
$\alpha = 0.0081$
207.
Logistic: $361,$ Threshold: $436,$ Gompertz: $309.$
Section 4.5 Exercises
209.
Yes
211.
Yes
213.
$y\prime - x^{3}y = \text{sin}\mspace{2mu} x$
215.
$y\prime + \frac{\left( {3x + 2} \right)}{x}y = \text{−}e^{x}$
217.
$\frac{dy}{dt} - yx\left( {x + 1} \right) = 0$
219.
$e^{(e^{x})}$
221.
$\text{−}\text{ln}\left( {\text{cosh}\mspace{2mu} x} \right)$
223.
$y = Ce^{3x} - \frac{2}{3}$
225.
$y = Cx^{3} + 6x^{2}$
227.
$y = Ce^{x^{2}\text{/}2} - 3$
229.
$y = C\mspace{2mu}\text{tan}\left( \frac{x}{2} \right) - 2x + 4\mspace{2mu}\text{tan}\left( \frac{x}{2} \right)\text{ln}\left( {\text{sin}\left( \frac{x}{2} \right)} \right)$
231.
$y = Cx^{3} - x^{2}$
233.
$y = C{(x + 2)}^{2} + \frac{1}{2}$
235.
$y = \frac{C}{\sqrt{x}} + 2\mspace{2mu}\text{sin}(3t)$
237.
$y = C{(x + 1)}^{3} - x^{2} - 2x - 1$
239.
$y = Ce^{\text{sinh}^{-1}x} - 2$
241.
$y = x + 4e^{–x} - 1$
243.
$y = - \frac{3x}{2}\left( {x^{2} - 1} \right)$
245.
$y = 1 - e^{\text{tan}^{-1}x}$
247.
$y = (x + 2)\text{ln}\mspace{2mu}\left( \frac{x + 2}{2} \right)$
249.
$y = 2e^{2\sqrt{x}} - 2x - 2\sqrt{x} - 1$
251.
$v(t) = \frac{gm}{k}\left( {1 - e^{\text{−}{{kt}\text{/}m}}} \right)$
253.
$40.451$ seconds
255.
$\sqrt{\frac{gm}{k}}$
257.
$y = Ce^{x} - a(x + 1)$
259.
$y = Ce^{x^{2}\text{/}2} - a$
261.
$y = \frac{e^{kt} - e^{t}}{k - 1}$
Review Exercises
263.
F
265.
T
267.
$y(x) = \frac{2^{x}}{\text{ln}(2)} + x\mspace{2mu}\text{cos}^{-1}x - \sqrt{1 - x^{2}} + C$
269.
$y(x) = \text{ln}\left( {C - \text{cos}\mspace{2mu} x} \right)$
271.
$y(x) = e^{e^{C + x}}$
273.
$y(x) = 4 + \frac{3}{2}x^{2} + 2x - \text{sin}\mspace{2mu} x$
275.
$y(x) = - \frac{2}{1 + 3\left( {x^{2} + 2\mspace{2mu}\text{sin}\mspace{2mu} x} \right)}$
277.
$y(x) = -2x^{2} - 2x - \frac{1}{3} - \frac{2}{3}e^{3x}$
279.
$y(x) = Ce^{\text{−}x} + \text{ln}\mspace{2mu} x$
281.
Euler: $0.6939,$ exact solution: $y(x) = \frac{3^{x} - e^{-2x}}{2 + \text{ln}(3)}$
283.
$\frac{40}{49}$ second
285.
$x(t) = 5000 + \frac{245}{9} - \frac{49}{3}t - \frac{245}{9}e^{\text{−}{5\text{/}{3t}}},t = 307.8$ seconds
287.
$T(t) = 200\left( {1 - e^{\text{−}{t\text{/}1000}}} \right)$
289.
$P(t) = \frac{1600000e^{0.02t}}{9840 + 160e^{0.02t}}$
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- Book title: Calculus Volume 2
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- Location: Houston, Texas
- Book URL: https://openstax.org/books/calculus-volume-2/pages/1-introduction
- Section URL: https://openstax.org/books/calculus-volume-2/pages/chapter-4
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