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14.6: Section 2.3 Answers

( \newcommand{\kernel}{\mathrm{null}\,}\)

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.

  1. all (x_{0},y_{0})
  2. (x_{0},y_{0}) woth y_{0}\neq 0

6. (b) all (x_{0}, y_{0})

7. (b) all (x_{0}, y_{0})

8. (b) (x_{0}, y_{0}) such that x_{0}\neq 4y_{0}

9.

  1. all (x_{0}, y_{0})
  2. all (x_{0}, y_{0})\neq (0,0)

10.

  1. all (x_{0}, y_{0})
  2. all (x_{0}, y_{0}) with y_{0}\neq\pm 1

11. (b) all (x_{0}, y_{0})

12. (b) all (x_{0}, y_{0}) such that (x_{0}, y_{0}) >0

13. (b) all (x_{0}, y_{0}) with 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, is a solution.

Also, y=\left\{\begin{array}{cc}{0,}&{-\infty <x\leq -\frac{5}{3}}\\[4pt]{(\frac{3}{5}x+1)^{5/3},}&{-\frac{5}{3}<x<\infty }\end{array} \right.\nonumber is a solution. For every a\geq \frac{5}{3}, the following function is also a solution: y=\left\{\begin{array}{cc}{(\frac{3}{5}(x+a))^{5/3},}&{-\infty <x<-a,}\\[4pt]{0,}&{-a\leq x\leq -\frac{5}{3}}\\[4pt]{(\frac{3}{5}x+1)^{5/3},}&{-\frac{5}{3}<x<\infty }\end{array} \right.\nonumber

17.

  1. all (x_{0}, y_{0})
  2. all (x_{0}, y_{0}) with 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,}\\[4pt]{1,}&{x<0}\end{array} \right.;\quad y_{7}=\left\{\begin{array}{cc}{1-x^{3},}&{x\geq 0,}\\[4pt]{1,}&{x<0}\end{array} \right.;\nonumber

y_{8}=\left\{\begin{array}{cc}{1,}&{x\geq 0,}\\[4pt]{1+x^{3},}&{x<0}\end{array} \right.;\quad y_{9}=\left\{\begin{array}{cc}{1,}&{x\geq 0,}\\[4pt]{1-x^{3},}&{x<0}\end{array} \right.\nonumber

19. y=1+(x^{2}+4)^{3/2},\quad -\infty <x<\infty

20.

  1. The solution is unique on (0,\infty ). It is given by
    y=\left\{\begin{array}{cc}{1,}&{0<x\leq \sqrt{5}}\\[4pt]{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty }\end{array} \right.\nonumber
  2. y=\left\{\begin{array}{cc}{1,}&{-\infty <x\leq\sqrt{5},}\\[4pt]{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty }\end{array} \right.\nonumber is a solution of (A) on (-\infty ,\infty ). If \alpha\geq 0, then y=\left\{\begin{array}{cc}{1+(x^{2}-\alpha ^{2})^{3/2},}&{-\infty <x<-\alpha ,}\\[4pt]{1,}&{-\alpha\leq x\leq\sqrt{5},}\\[4pt]{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty ,}\end{array}\right.\nonumber and y=\left\{\begin{array}{cc}{1-(x^{2}-\alpha ^{2})^{3/2},}&{-\infty <x<-\alpha ,}\\[4pt]{1,}&{-\alpha\leq x\leq\sqrt{5},}\\[4pt]{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty ,}\end{array} \right.\nonumber are also solutions of (A) on (-\infty ,\infty).

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