14.22: Sección 2.3 Respuestas
- Page ID
- 115205
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)1. b)\(x_{0}\neq k\pi \;\;\;\;(k=\text{integer})\)
2. b)\( (x_{0},y_{0})\neq (0,0)\)
3. b)\(x_{0}y_{0}\neq (2k+1)\frac{\pi }{2} \;\;\;\; (k=\text{integer})\)
4. b)\(x_{0}y_{0}>0\text{ and }x_{0}y_{0}\neq 1\)
5.
- todos\((x_{0},y_{0})\)
- \((x_{0},y_{0})\)woth\(y_{0}\neq 0\)
6. b) todos\((x_{0}, y_{0})\)
7. b) todos\((x_{0}, y_{0})\)
8. b) de\((x_{0}, y_{0})\) tal manera que\(x_{0}\neq 4y_{0}\)
9.
- todos\((x_{0}, y_{0})\)
- todos\((x_{0}, y_{0})\neq (0,0)\)
10.
- todos\((x_{0}, y_{0})\)
- todo\((x_{0}, y_{0})\) con\(y_{0}\neq\pm 1\)
11. b) todos\((x_{0}, y_{0})\)
12. b) todos\((x_{0}, y_{0})\) tales que\((x_{0}, y_{0}) >0\)
13. b) todos\((x_{0}, y_{0})\) con\(x_{0}\neq 1,\quad y_{0}\neq (2k+1)\frac{\pi }{2}(k=\text{integer})\)
16. \(y=\left(\frac{3}{5}x+1 \right)^{5/3},\quad -\infty <x<\infty\), es una solución.
También,\[y=\left\{\begin{array}{cc}{0,}&{-\infty <x\leq -\frac{5}{3}}\\{(\frac{3}{5}x+1)^{5/3},}&{-\frac{5}{3}<x<\infty }\end{array} \right.\nonumber\] es una solución. Para cada uno\(a\geq \frac{5}{3}\), la siguiente función es también una solución:\[y=\left\{\begin{array}{cc}{(\frac{3}{5}(x+a))^{5/3},}&{-\infty <x<-a,}\\{0,}&{-a\leq x\leq -\frac{5}{3}}\\{(\frac{3}{5}x+1)^{5/3},}&{-\frac{5}{3}<x<\infty }\end{array} \right.\nonumber\]
17.
- todos\((x_{0}, y_{0})\)
- todo\((x_{0}, y_{0})\) con\(y_{0}\neq 1\)
18. \(y_{1}=1; y_{2}=1+|x|^{3};y_{3}=1-|x|^{3};y_{4}=1+x^{3};y_{5}=1-x^{3}\)
\[y_{6}=\left\{\begin{array}{cc}{1+x^{3},}&{x\geq 0,}\\{1,}&{x<0}\end{array} \right.;\quad y_{7}=\left\{\begin{array}{cc}{1-x^{3},}&{x\geq 0,}\\{1,}&{x<0}\end{array} \right.;\nonumber\]
\[y_{8}=\left\{\begin{array}{cc}{1,}&{x\geq 0,}\\{1+x^{3},}&{x<0}\end{array} \right.;\quad y_{9}=\left\{\begin{array}{cc}{1,}&{x\geq 0,}\\{1-x^{3},}&{x<0}\end{array} \right.\nonumber\]
19. \(y=1+(x^{2}+4)^{3/2},\quad -\infty <x<\infty \)
20.
- La solución es única en\((0,\infty )\). Está dado por
\[y=\left\{\begin{array}{cc}{1,}&{0<x\leq \sqrt{5}}\\{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty }\end{array} \right.\nonumber\] - \[y=\left\{\begin{array}{cc}{1,}&{-\infty <x\leq\sqrt{5},}\\{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty }\end{array} \right.\nonumber\]es una solución de (A) on\((-\infty ,\infty )\). Si\(\alpha\geq 0\), entonces\[y=\left\{\begin{array}{cc}{1+(x^{2}-\alpha ^{2})^{3/2},}&{-\infty <x<-\alpha ,}\\{1,}&{-\alpha\leq x\leq\sqrt{5},}\\{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty ,}\end{array}\right.\nonumber\] y también\[y=\left\{\begin{array}{cc}{1-(x^{2}-\alpha ^{2})^{3/2},}&{-\infty <x<-\alpha ,}\\{1,}&{-\alpha\leq x\leq\sqrt{5},}\\{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty ,}\end{array} \right.\nonumber\] son soluciones de (A) en\((-\infty ,\infty)\).