User Tools


no way to compare when less than two revisions

Differences

This shows you the differences between two versions of the page.


de2de:chapter-6 [2012/06/25 11:33] (current) – created dcs
Line 1: Line 1:
 +====Differential Equations to Difference Equations====
  
 +===Answers for selected problems===
 +
 +**Chapter 6**
 +
 +----
 +
 +**Section 6.1**
 +
 +1. (a) $ f'(x) = 6e^{2x} $ 
 +
 +%%(c)%% $ h'(z) = 3z(15z^3 - 30z + 2)e^{5z^3} $ 
 +
 +(e) $ \displaystyle{g'(z) = \frac{3}{2}(1 - x)e^{-x}}$ 
 +
 +(g) $ \displaystyle{f'(s) = \frac{(1 + 6s)e^{-2s} + 6}{(e^{-2s} + 2)^2}}$ 
 +
 +
 +2. (a) $ \displaystyle{\int 3e^{2x}dx = \frac{3}{2}e^{2x} + c}$ 
 +
 +%%(c)%% $ \displaystyle{\int 4te^{3t}dt = \frac{4}{3}te^{3t} - \frac{4}{9}e^{3t} + c}$ 
 +
 +(e) $ \displaystyle{\int z^2 e^z dz = z^2e^z - 2ze^z + 2e^z + c}$ 
 +
 +(g) $ \displaystyle{\int e^x\cos(x)dx = \frac{1}{2}(\cos(x) + \sin(x))e^x + c}$ 
 +
 +
 +3. The maximum value of $ f $ is $ 4e^{-2} $ at $ x = 2 $. 
 +
 +
 +5. (a) $ \displaystyle{\lim_{x \to \infty}xe^{-x} = 0}$ 
 +
 +%%(c)%% $ \displaystyle{\lim_{t \to 0}\frac{e^{-t} - 1}{t} = -1}$ 
 +
 +
 +8. (a) $ \displaystyle{\int_0^\infty e^{-x}dx = 1}$ 
 +
 +%%(c)%% $ \displaystyle{\int_0^\infty xe^{-x}dx = 1}$ 
 +
 +(e) $ \displaystyle{\int_0^\infty x^2e^{-x}dx = 2}$ 
 +
 +
 +10. (a) For $ -\infty < x < \infty $,
 +\[
 +e^{-x^2} = \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{n!} = 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{3!} + \frac{x^8}{4!} - \cdots .
 +\]
 +
 +%%(c)%% For $ -\infty < x < \infty $,
 +\[
 +\mathrm{erf}(x) = \frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)n!} = \frac{2x}{\sqrt{\pi}} - \frac{2x^3}{3\sqrt{\pi}} + \frac{x^5}{5\sqrt{\pi}} - \frac{x^7}{21\sqrt{\pi})} + \frac{x^9}{108\sqrt{\pi}} - \cdots .
 +\]
 +
 +%%(c)%% $ \mathrm{erf}(1) \approx P_{13}(1) = 0.84271 $, rounded to 5 decimal places 
 +
 +
 +11. (a) $ 26.4 $ million, rounded to one decimal place 
 +
 +(b) $ 2072 $
 +
 +----
 +
 +**Section 6.2**
 +
 +1. (a) $ \log(6) = a + b $ 
 +
 +%%(c)%% $ \log(9) = 2b $ 
 +
 +
 +2. (a) $ \displaystyle{f'(x) = \frac{2}{x}}$ 
 +
 +%%(c)%% $ \displaystyle{g'(x) = \frac{2}{x} + \frac{x}{x^2 + 5}}$ 
 +
 +(e) $ \displaystyle{f'(x) = \frac{e^{2x}}{x} + 2e^{2x}\log(5x)}$ 
 +
 +(g) $ \displaystyle{h'(x) = \frac{1}{x\log(x)}}$ 
 +
 +(i) $ \displaystyle{f'(x) = ex^{e-1}}$ 
 +
 +
 +3. (a) $ \displaystyle{\int \frac{1}{2x} \ dx = \frac{1}{2}\log|x| + c}$ 
 +
 +%%(c)%% $ \displaystyle{\int \frac{5x}{3x^2 + 1} \ dx = \frac{5}{6}\log(3x^2 + 1) + c}$ 
 +
 +(e) $ \displaystyle{\int \tan(3x)dx = -\frac{1}{3}\log|\cos(3x)| + c}$ 
 +
 +(g) $ \displaystyle{\int \csc(x)dx = -\log|\csc(x) + \cot(x)| + c}$ 
 +
 +
 +4. (a) $ \displaystyle{\int \log(3x)dx = x\log(3x) - x + c}$ 
 +
 +%%(c)%% $ \displaystyle{\int \frac{\log(x)}{x} \ dx = \frac{1}{2}(\log(x))^2 + c}$ 
 +
 +(e) $ \displaystyle{\int \log(x + 1)dx = (x + 1)\log(x + 1) - x + c}$ 
 +
 +(g) $ \displaystyle{\int \frac{1}{x\log(x)} \ dx = \log|\log(x)| + c}$ 
 +
 +
 +6. (b) $ \log(2) \approx P_{200}(2) = 0.6907 $, rounded to 4 decimal places 
 +
 +%%(c)%% $ \log(1.5) \approx P_7(1.5) = 0.4058 $, rounded to 4 decimal places 
 +
 +
 +10. (a) $ \displaystyle{\lim_{x \to \infty}\log(\log(x)) = \infty}$ 
 +
 +(d) $ x = e^{20} =  485,165,195.4 $, rounded to one decimal place 
 +
 +(e) $ x = e^{\left(e^{20}\right)} = 1.50655428 \times 10^{210,704,567} $, rounded to 8 decimal places 
 +
 +
 +11. Length $ \displaystyle{= \int_1^{10} \sqrt{1 + \frac{1}{x^2}} \ dx = 9.4172}$, rounded to four decimal places 
 +
 +
 +12. $ \displaystyle{x(t)  = 100e^{\alpha t}}$, where $ \alpha =\dfrac{\log(2)}{5} $
 +
 +----
 +
 +**Section 6.3**
 +
 +1. (a) $ \displaystyle{\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e}$ 
 +
 +%%(c)%% $ \displaystyle{\lim_{n \to \infty}\left(1 - \frac{2}{n}\right) = e^{-2}}$ 
 +
 +(e) $ \displaystyle{\lim_{n \to \infty}\left(1 + \frac{2}{n^2}\right) = 1}$ 
 +
 +(f) $ \displaystyle{\lim_{n \to \infty}\left(1 - \frac{4}{n} + \frac{1}{n^2}\right) = e^{-4}}$ 
 +
 +
 +3. (a) $\$6946.20$ 
 +
 +
 +(b) $ %%$%%6961.83 $ 
 +
 +%%(c)%% $ %%$%%6967.91 $ 
 +
 +(d) $ %%$%%6969.48 $ 
 +
 +(e) $ %%$%%6969.74 $ 
 +
 +
 +4. The $ 5.5 $% interest compounded quarterly is the most advantageous.
 +
 +6. $ 24,765 $ years 
 +
 +
 +7. The bone is between $ 23,257 $ years and $ 26,609 $ years old. 
 +
 +
 +8. $ x(t) = x_0e^{-0.034657t} $
 +$ 66.4 $ minutes until 10% remains; $ 86.4 $ minutes until 5% remains 
 +
 +
 +9. $ x(t) = x_0e^{-0.000028881t} $
 +$ 79,726 $ years until 10% remains; $ 103,726 $ years until 5% remains 
 +
 +
 +10. $ 69 $ years 
 +
 +
 +11. (a) $ y(t) = 179.3e^{0.012562t} $ 
 +
 +(b) Prediction for 1980: $ 230.5 $ million
 +
 +Prediction for 1990: $ 261.4 $ million
 +
 +Prediction for 2000: $ 296.3 $ million 
 +
 +13. (a) $ x(t) = 75.995e^{0.0190825t} $, in millions
 +
 +Prediction for 1920: $ 111.3 $ million
 +
 +Prediction for 1930: $ 134.7 $ million
 +
 +Prediction for 1950: $ 197.3 $ million
 +
 +Prediction for 1970: $ 289.0 $ million
 +
 +Prediction for 1990: $ 423.3 $ million
 +
 +Prediction for 2000: $ 512.3 $ million
 +
 +The population in 1936 would be twice the population of 1900.
 +
 +(b) $ \displaystyle{x(t) = \frac{10595}{75.995 + 63.424e^{-0.048106t}}}$, in millions
 +
 +Prediction for 1930: $ 116.5 $ million
 +
 +Prediction for 1950: $ 129.7 $ million
 +
 +Prediction for 1970: $ 135.5 $ million
 +
 +Prediction for 1990: $ 137.9 $ million
 +
 +Prediction for 2000: $ 138.5 $ million
 +
 +This model predicts a limiting population of $ M = 139.4 $ million. Since this is less than twice the population in 1900, this model predicts that the population will never double from its 1900 level. 
 +
 +
 +14. Hint: $ \displaystyle{\ddot{x}(t) = \alpha \dot{x}(t) - \frac{2\alpha}{M}x(t)\dot{x}(t)}$
 +
 +This tells us that the rate of growth of the population is increasing when the population is less than one-half of its limiting size and decreasing when the population exceeds one-half of its limiting size.
 +
 +----
 +
 +**Section 6.4**
 +
 +1. (a) $ \displaystyle{\int \frac{x-1}{x} \ dx = x - \log|x| + c}$ 
 +
 +%%(c)%% $ \displaystyle{\int \frac{3x^2}{x-2} \ dx = \frac{3}{2}x^2 + 6x + 12\log|x-2| + c}$ 
 +
 +(e) $ \displaystyle{\int \frac{4x+1}{2x^2+x-3} \ dx = \log|2x^2 + x - 3| + c}$ 
 +
 +
 +2. (a) $ \displaystyle{\int \frac{1}{(x+2)(x-4)} \ dx = \frac{1}{6}\log|x-4| - \frac{1}{6}\log|x+2| + c}$ 
 +
 +%%(c)%% $ \displaystyle{\int \frac{3x}{(2x+3)(x+1)} \ dx = \frac{9}{2}\log|2x+3| - 3\log|x+1| + c}$ 
 +
 +(e) $ \displaystyle{\int \frac{x}{x^2+x-6} \ dx = \frac{2}{5}\log|x-2| + \frac{3}{5}\log|x+3| + c}$ 
 +
 +(g) $ \displaystyle{\int \frac{3}{x^2+5x+6} \ dx = 3\log|x+2| - 3\log|x+3| + c}$ 
 +
 +
 +3. (a) $ \displaystyle{\frac{1}{(x-1)^2} \ dx = -\frac{1}{x-1} + c}$ 
 +
 +%%(c)%% $ \displaystyle{\int \frac{x}{x^2+2x+1} \ dx = \frac{1}{x+1} + \log|x+1| + c}$ 
 +
 +(e) $ \displaystyle{\int \frac{5}{(x+2)^3} \ dx = -\frac{5}{2(x+2)^2} + c}$ 
 +
 +(g) $ \displaystyle{\int \frac{3x^2}{(x+1)^2(x-3)} \ dx = \frac{3}{4(x+1)} + \frac{21}{16}\log|x+1| + \frac{27}{16}\log|x-3| + c}$ 
 +
 +
 +4. (a) $ \displaystyle{\int \frac{1}{(3x+2)^2} \ dx = -\frac{1}{3(3x+2)} + c}$ 
 +
 +%%(c)%% $ \displaystyle{\int \frac{9x^2-4x}{3x^3-2x^2+5} \ dx = \log|3x^3 - 2x^2 + 5| + c}$ 
 +
 +(e) $ \displaystyle{\int_{-1}^1 \frac{1}{x^2-4} \ dx = -\frac{1}{2}\log(3)}$ 
 +
 +(g) $ \displaystyle{\int \frac{4x+5}{(x-2)^2(x+5)} \ dx = -\frac{13}{7(x-2)} + \frac{15}{49}\log|x-2| - \frac{15}{49}\log|x+5| + c}$ 
 +
 +
 +5. $ \displaystyle{x(t) = \frac{1-e^{2t}}{1+e^{2t}}}$
 +
 +----
 +
 +**Section 6.5**
 +
 +1. (a) $ \displaystyle{f'(x) = \frac{x}{1+x^2} + \tan^{-1}(x)}$ 
 +
 +%%(c)%% $ \displaystyle{g'(x) = \frac{3}{x\sqrt{1-9x^2}} - \frac{\sin^{-1}(3x)}{x^2}}$ 
 +
 +
 +2. (a) $ \displaystyle{\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}}$ 
 +
 +%%(c)%% $ \displaystyle{\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}}$ 
 +
 +(e) $ \displaystyle{\sin^{-1}\left(\sin\left(\frac{3\pi}{4}\right)\right) = \frac{\pi}{4}}$ 
 +
 +(f) $ \displaystyle{\sin\left(\sin^{-1}\left(-\frac{1}{\sqrt{2}}\right)\right) = -\frac{1}{\sqrt{2}}}$ 
 +
 +3. (a) $ \displaystyle{\int \frac{1}{1 + 2x^2} \ dx = \frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}x) + c}$ 
 +
 +%%(c)%% $ \displaystyle{\int \frac{3}{x^2 + 4} \ dx = \frac{3}{2}\tan^{-1}\left(\frac{x}{2}\right) + c}$ 
 +
 +(e) $ \displaystyle{\int \frac{x}{x^2+4x+5} \ dx = \frac{1}{2}\log(x^2 + 4x + 5) - 2\tan^{-1}(x + 2) + c}$ 
 +
 +(g) $ \displaystyle{\int_{-1}^1 \frac{1}{1 + x^2} \ dx = \frac{\pi}{2}}$ 
 +
 +
 +4. (a) $ \displaystyle{\int \frac{1}{x^3+x} \ dx = \log|x| - \frac{1}{2}\log(1 + x^2) + c}$ 
 +
 +%%(c)%% $ \displaystyle{\int \frac{1}{x^2(x^2+1)} \ dx = -\frac{1}{x} - \tan^{-1}(x) + c}$ 
 +
 +(e) $ \displaystyle{\int \sin^{-1}(x)dx = x\sin^{-1}(x) + \sqrt{1 - x^2} + c}$ 
 +
 +(g) $ \displaystyle{\int \frac{5}{\sqrt{1-9x^2}} \ dx = \frac{5}{3}\sin^{-1}(3x) + c}$ 
 +
 +(i) $ \displaystyle{\int \frac{3x}{\sqrt{1-x^2}} \ dx = -3\sqrt{1-x^2} + c}$ 
 +
 +
 +8. $ \displaystyle{\int_{-\infty}^\infty \frac{1}{1 + x^2} \ dx = \pi}$ 
 +
 +
 +9. (a) For $ -1 \le x \le 1 $, 
 +\[
 +\tan^{-1}(x) = \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2n + 1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \cdots .
 +\]
 +
 +(b) $ \displaystyle{\pi = 4\tan^{-1}(1) \approx P_{3999}(1) = 3.1411}$, rounded to 4 decimal places
 +
 +----
 +
 +**Section 6.6**
 +
 +1. (a) $ \displaystyle{\int \frac{3}{\sqrt{16-x^2}} \ dx = 3\sin^{-1}\left(\frac{x}{4}\right) + c}$ 
 +
 +%%(c)%% $ \displaystyle{\int \sqrt{5-z^2} \ dz = \frac{z}{2}\sqrt{5-z^2} + \frac{5}{2}\sin^{-1}\left(\frac{z}{\sqrt{5}}\right) + c}$ 
 +
 +(e) $ \displaystyle{\int \frac{1}{\sqrt{4+x^2}} \ dx = \log\left|\sqrt{4+x^2} + x\right| + c}$ 
 +
 +
 +2. (a) $ \displaystyle{\int z\sqrt{1-z^2} \ dz = -\frac{1}{3}(1 - z^2)^{\frac{3}{2}} + c}$ 
 +
 +%%(c)%% $ \displaystyle{\int \frac{1}{\sqrt{x^2 - 4}} \ dx = \log\left|x + \sqrt{x^2-4}\right| + c}$ 
 +
 +(e) $ \displaystyle{\int \frac{4}{\sqrt{3-2x^2}}  \ dx = 2\sqrt{2}\sin^{-1}\left(\sqrt{\frac{2}{3}}x\right) + c}$ 
 +
 +(g) $ \displaystyle{\int \sqrt{4-t^2} \ dt = \frac{t}{2}\sqrt{4-t^2} + 2\sin^{-1}\left(\frac{t}{2}\right)}$ 
 +
 +
 +3. (a) $ \displaystyle{\int_0^3 \sqrt{9-t^2} \ dt = \frac{9\pi}{4}}$ 
 +
 +%%(c)%% $ \displaystyle{\int_0^1 \frac{1}{(1+x^2)^{\frac{3}{2}}} \ dx = \frac{1}{\sqrt{2}}}$ 
 +
 +(e) $ \displaystyle{\int_{5\sqrt{2}}^{10} \frac{1}{\sqrt{x^2-25}} \ dx = \log(2+\sqrt{3}) - \log(1 + \sqrt{2})}$ 
 +
 +
 +5. $ \displaystyle{\int \frac{1}{\sqrt{1-x^2}} \ dx = -\cos^{-1}(x) + c}$
 +
 +----
 +
 +**Section 6.7**
 +
 +1. (a) $ f'(x) = 3\cosh(3x) $ 
 +
 +%%(c)%% $ f'(t) = 6t\sinh(t)\sinh(2t) + 3t\cosh(t)\cosh(2t) + 3\sinh(t)\cosh(t) $ 
 +
 +(e) $ y'(t) = 40t^2\cosh(4t)\sinh(4t) + 10t\cosh^2(4t) $ 
 +
 +
 +2. (a) $ \displaystyle{\int \sinh(3x)dx = \frac{1}{3}\cosh(3x) + c}$ 
 +
 +%%(c)%% $ \displaystyle{\int \sinh(z)\cosh(z)dz = \frac{1}{2}\sinh^2(z) + c}$ 
 +
 +(e) $ \displaystyle{\int e^{-2t}\cosh(2t)dt = \frac{1}{2}t - \frac{1}{8}e^{-4t} + c}$ 
 +
 +(g) $ \displaystyle{\int 5t^2\cosh(2t)dt = \frac{5}{2}t^2\sinh(2t) - \frac{5}{2}t\cosh(2t) + \frac{5}{4}\sinh(2t) + c}$ 
 +
 +
 +4. (a) $ \displaystyle{\int \frac{1}{\sqrt{4+x^2}} \ dx = \sinh^{-1}\left(\frac{x}{2}\right) + c}$ 
 +
 +%%(c)%% $ \displaystyle{\int \frac{3}{\sqrt{9+3t^2}} \ dt = \sqrt{3}\sinh^{-1}\left(\frac{t}{\sqrt{3}}\right) + c}$ 
 +
 +(d) For $ x < -1 $, $ \displaystyle{\int \frac{1}{\sqrt{x^2-1}} \ dx = -\cosh^{-1}|x| + c}$. 
 +
 +
 +6. (a) $ \displaystyle{f'(x) = 12x\mathrm{sech}^2(4x) + 3\tanh(4x)}$ 
 +
 +%%(c)%% $ \displaystyle{h'(\theta) = 8\tanh(\theta)\mathrm{sech}^3(\theta) - 4\tanh^3(\theta)\mathrm{sech}(\theta)}$ 
 +
 +
 +7. (a) $ \displaystyle{\int \tanh(x)dx = \log(\cosh(x)) + c}$ 
 +
 +%%(c)%% $ \displaystyle{\int \frac{1}{4 - x^2} \ dx = \frac{1}{2}\tanh^{-1}\left(\frac{x}{2}\right) + c}$
 +
 +----