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11.19: A.2.2- Section 2.2 Answers

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Dec 31, 2022
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- 11.18: A.2.1- Section 2.1 Answers
- 11.20: A.2.3- Section 2.3 Answers

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1. $y=2\pm\sqrt{2(x^{3}+x^{2}+x+c)}$

2. $\ln (|\sin y|)=\cos x+c; \quad y ≡ k\pi,\quad k=\text{integer}$

3. $y=\frac{c}{x-c}\quad y≡-1$

4. $\frac{(\ln y)^{2}}{2}=-\frac{x^{3}}{3}+c$

5. $y^{3}+3\sin y+\ln |y|+\ln (1+x^{2})+\tan ^{-1}x=c;\quad y≡0$

6. $y=\pm\left( 1+\left(\frac{x}{1+cx}\right) ^{2}\right) ^{1/2} ;\quad y≡\pm 1$

7. $y=\tan \left(\frac{x^{3}}{3}+c \right)$

8. $y=\frac{c}{\sqrt{1+x^{2}}}$

9. $y=\frac{2-ce^{(x-1)^{2}/2}}{1-ce^{(x-1)^{2}/2}};\quad y≡1$

10. $y=1+(3x^{2}+9x+c)^{1/3}$

11. $y=2+\sqrt{\frac{2}{3}x^{3}+3x^{2}+4x-\frac{11}{3}}$

12. $y=\frac{e^{-(x^{2}-4)/2}}{2-e^{-(x^{2}-4)/2}}$

13. $y^{3}+2y^{2}+x^{2}+\sin x=3$

14. $(y+1)(y-1)^{-3}(y-2)^{2}=-256(x+1)^{-6}$

15. $y=-1+3e^{-x^{2}}$

16. $y=\frac{1}{\sqrt{2e^{-2x^{2}}-1}}$

17. $y≡-1;\quad (-\infty ,\infty )$

18. $y=\frac{4-e^{-x^{2}}}{2-e^{-x^{2}}}; \quad (-\infty ,\infty )$

19. $y=\frac{-1+\sqrt{4x^{2}-15}}{2}; \quad \left(\frac{\sqrt{15}}{2},\infty \right)$

20. $y=\frac{2}{1+e^{-2x}}\quad (-\infty ,\infty )$

21. $y=-\sqrt{25-x^{2}};\quad (-5,5)$

22. $y≡2,\quad (-\infty ,\infty )$

23. $y=3\left(\frac{x+1}{2x-4} \right)^{1/3};\quad (-\infty ,2)$

24. $y=\frac{x+c}{1-cx}$

25. $y=-x\cos c+\sqrt{1-x^{2}}\sin c;\quad y≡1; \:y≡-1$

26. $y=-x+3\pi /2$

28. $P=\frac{P_{0}}{\alpha P_{0}+(1-\alpha P_{0})e^{-at}};\lim_{t\to\infty }P(t)=1/\alpha$

29. $I=\frac{SI_{0}}{I_{0}+(S-I_{0})e^{-rSt}}$

30. $\text{If }q=rS\text{ then }I=\frac{I_{0}}{1+rI_{0}t}\text{ and }\lim_{t\to\infty }I(t)=0.\text{ If }q\neq Rs\text{, then }I=\frac{\alpha I_{0}}{I_{0}+(\alpha -I_{0})e^{-r\alpha t}}.\text{ If } q<rs\text{, then }\lim_{t\to\infty}I(t)=\alpha =S-\frac{q}{r}\text{ if }q>rS\text{, then }\lim_{t\to\infty}I(t)=0$

34. $f=ap,\quad \text{where }a=\text{constant}$

35. $y=e^{-x}(-1\pm\sqrt{2x^{2}+c})$

36. $y=x^{2}(-1+\sqrt{x^{2}+c})$

37. $y=e^{x}(-1+(3xe^{x}+c)^{1/3})$

38. $y=e^{2x}(1\pm\sqrt{c-x^{2}})$

39.

$y_{1}=1/x;\quad g(x)=h(x)$
$y_{1}=x;\quad g(x)=h(x)/x^{2}$
$y_{1}=e^{-x};\quad g(x)=e^{x}h(x)$
$y_{1}=x^{-r};\quad g(x)=x^{r-1}h(x)$
$y_{1}=1/v(x);\quad g(x)=v(x)h(x)$

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