A.2.3: Section 2.3 Answers
( \newcommand{\kernel}{\mathrm{null}\,}\)
1. (b) x0≠kπ(k=integer)
2. (b) (x0,y0)≠(0,0)
3. (b) x0y0≠(2k+1)π2(k=integer)
4. (b) x0y0>0 and x0y0≠1
5.
- all (x0,y0)
- (x0,y0) woth y0≠0
6. (b) all (x0,y0)
7. (b) all (x0,y0)
8. (b) (x0,y0) such that x0≠4y0
9.
- all (x0,y0)
- all (x0,y0)≠(0,0)
10.
- all (x0,y0)
- all (x0,y0) with y0≠±1
11. (b) all (x0,y0)
12. (b) all (x0,y0) such that (x0,y0)>0
13. (b) all (x0,y0) with x0≠1,y0≠(2k+1)π2(k=integer)
16. y=(35x+1)5/3,−∞<x<∞, is a solution.
Also, y={0,−∞<x≤−53(35x+1)5/3,−53<x<∞ is a solution. For every a≥53, the following function is also a solution: y={(35(x+a))5/3,−∞<x<−a,0,−a≤x≤−53(35x+1)5/3,−53<x<∞
17.
- all (x0,y0)
- all (x0,y0) with y0≠1
18. y1=1;y2=1+|x|3;y3=1−|x|3;y4=1+x3;y5=1−x3
y6={1+x3,x≥0,1,x<0;y7={1−x3,x≥0,1,x<0;
y8={1,x≥0,1+x3,x<0;y9={1,x≥0,1−x3,x<0
19. y=1+(x2+4)3/2,−∞<x<∞
20.
- The solution is unique on (0,∞). It is given by
y={1,0<x≤√51−(x2−5)3/2,√5<x<∞ - y={1,−∞<x≤√5,1−(x2−5)3/2,√5<x<∞ is a solution of (A) on (−∞,∞). If α≥0, then y={1+(x2−α2)3/2,−∞<x<−α,1,−α≤x≤√5,1−(x2−5)3/2,√5<x<∞, and y={1−(x2−α2)3/2,−∞<x<−α,1,−α≤x≤√5,1−(x2−5)3/2,√5<x<∞, are also solutions of (A) on (−∞,∞).