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11.20: A.2.3- Section 2.3 Answers

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Dec 31, 2022
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- 11.19: A.2.2- Section 2.2 Answers
- 11.21: A.2.4- Section 2.4 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) $x_{0}y_{0}>0\text{ and }x_{0}y_{0}\neq 1$

all $(x_{0},y_{0})$
$(x_{0},y_{0})$ 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 $x_{0}\neq 4y_{0}$

all $(x_{0}, y_{0})$
all $(x_{0}, y_{0})\neq (0,0)$

10.

all $(x_{0}, y_{0})$
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.

all $(x_{0}, y_{0})$
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.

The solution is unique on $(0,\infty )$ . It is given by
$y=\left\{\begin{array}{cc}{1,}&{0<x\leq \sqrt{5}}\\[4pt]{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty }\end{array} \right.\nonumber$
$y=\left\{\begin{array}{cc}{1,}&{-\infty <x\leq\sqrt{5},}\\[4pt]{1-(x^{2}-5)^{3/2},}&{\sqrt{5}<x<\infty }\end{array} \right.\nonumber$ 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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