# 11.20: A.2.3- Section 2.3 Answers

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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 $$(-\infty ,\infty)$$.

This page titled 11.20: A.2.3- Section 2.3 Answers is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.