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14.47: Section 10.1 Answers

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Aug 11, 2020
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- 14.46: Section 9.4 Answers
- 14.48: Section 10.2 Answers

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1. $\begin{array}{l}{Q'_{1}=2-\frac{1}{10}Q_{1}+\frac{1}{25}Q_{2}}\\[4pt]{Q'_{2}=6+\frac{3}{50}Q_{1}-\frac{1}{20}Q_{2}}\end{array}$

2. $\begin{array}{l}{Q'_{1}=12-\frac{5}{100+2t}Q_{1}+\frac{1}{100+3t}Q_{2}}\\[4pt]{Q'_{2}=5+\frac{1}{50+t}Q_{1}-\frac{4}{100+3t}Q_{2}}\end{array}$

3. $\begin{array}{l}{m_{1}y_{1}''=-(c_{1}+c_{2})y'_{1}+c_{2}y'_{2}-(k_{1}+k_{2})y_{1}+k_{2}y_{2}+F_{1}}\\[4pt]{m_{2}y_{2}''=(c_{2}-c_{3})y'_{1}-(c_{2}+c_{3})y'_{2}+c_{3}y'_{3}+(k_{2}-k_{3})y_{1}-(k_{2}+k_{3})y_{2}+k_{3}y_{3}+F_{2}}\\[4pt]{m_{3}y_{3}''=c_{3}y'_{1}+c_{3}y'_{2}-c_{3}y'_{3}+k_{3}y_{1}+k_{3}y_{2}-k_{3}y_{3}+F_{3}}\end{array}$

4. $x''=-\frac{\alpha}{m}x'+\frac{gR^{2}x}{(x^{2}+y^{2}+x^{2})^{3/2}}\qquad y''=-\frac{\alpha}{m}y'+\frac{gR^{2}y}{(x^{2}+y^{2}+x^{2})^{3/2}}\qquad z''=-\frac{\alpha}{m}z'+\frac{gR^{2}z}{(x^{2}+y^{2}+z^{2})^{3/2}}$

$\begin{array}{l}{x'_{1}=x_{2}}\\[4pt]{x'_{2}=x_{3}}\\[4pt]{x'_{3}=f(t,x_{1},y_{1},y_{2})}\\[4pt]{y'_{1}=y_{2}}\\[4pt]{y'_{2}=g(t,y_{1},y_{2})}\end{array}$
$\begin{array}{l}{u'_{1}=f(t,u_{1},v_{1},v_{2},w_{2})}\\[4pt]{v'_{1}=v_{2}}\\[4pt]{v'_{2}=g(t,u_{1},v_{1},v_{2},w_{1})}\\[4pt]{w'_{1}=w_{2}}\\[4pt]{w'_{2}=h(t,u_{1},v_{1},v_{2},w_{1},w_{2})}\end{array}$
$\begin{array}{l}{y'_{1}=y_{2}}\\[4pt]{y'_{2}=y_{3}}\\[4pt]{y'_{3}=f(t,y_{1},y_{2},y_{3})}\end{array}$

$\begin{array}{l}{y'_{1}=y_{2}}\\[4pt]{y'_{2}=y_{3}}\\[4pt]{y'_{3}=y_{4}}\\[4pt]{y'_{4}=f(t,y_{1})}\end{array}$
$\begin{array}{l}{x'_{1}=x_{2}}\\[4pt]{x'_{2}=f(t,x_{1},y_{1})}\\[4pt]{y'_{1}=y_{2}}\\[4pt]{y'_{2}=g(t,x_{1},y_{1})}\end{array}$

6. $\begin{array}{ll}{x'=x_{1}}&{x'_{1}=-\frac{gR^{2}x}{(x^{2}+y^{2}+x^{2})^{3/2}}}\\[4pt]{y'=y_{1}}&{y'_{1}=-\frac{gR^{2}y}{(x^{2}+y^{2}+x^{2})^{3/2}}}\\[4pt]{z'=z_{1}}&{z'_{1}=-\frac{gR^{2}z}{(x^{2}+y^{2}+z^{2})^{3/2}}}\end{array}$

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