# 11.31: A.4.5- Section 4.5 Answers

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1. $$y'=\frac{2xy}{x^{2}+3y^{2}}$$

2. $$y'=-\frac{y^{2}}{(xy-1)}$$

3. $$y'=-\frac{y(x^{2}+y^{2}-2x^{2}\ln |xy|)}{x(x^{2}+y^{2}-2y^{2}\ln |xy|)}$$

4. $$xy'-y=-\frac{x^{1/2}}{2}$$

5. $$y'+2xy=4xe^{x^{2}}$$

6. $$xy'+y=4x^{3}$$

7. $$y' − y = \cos x − \sin x$$

8. $$(1+x^{2})y'-2xy=(1-x)^{2}e^{x}$$

10. $$y'g-yg'=f'g-fg'$$

11. $$(x-x_{0})y'=y-y_{0}$$

12. $$y'(y^{2}-x^{2}+1)+2xy=0$$

13. $$2x(y-1)y'-y^{2}+x^{2}+2y=0$$

14.

1. $$y=-81+18x,\: (9,81)\quad y=-1+2x,\: (1,1)$$
2. $$y = −121 + 22x,\: (11, 121)\quad y = −1 + 2x,\: (1, 1)$$
3. $$y = −100 − 20x,\: (−10, 100)\quad y = −4 − 4x,\: (−2, 4)$$
4. $$y = −25 − 10x,\: (−5, 25)\quad y = −1 − 2x,\: (−1, 1)$$

15. (e) $$y=\frac{5+3x}{4},\:(-3/5,4/5)\quad y=-\frac{5-4x}{3},\: (4/5,-3/5)$$

17.

1. $$y=-\frac{1}{2}(1+x),\: (1,-1);\quad y=\frac{5}{2}+\frac{x}{10},\: (25,5)$$
2. $$y=\frac{1}{4}(4+x),\: (4,2)\quad y=-\frac{1}{4}(4+x),\: (4,-2)$$
3. $$y = \frac{1}{2}(1 + x),\: (1, 1)\quad y = \frac{7}{2} + \frac{x}{14},\: (49, 7)$$
4. $$y = −\frac{1}{2}(1 + x),\: (1, −1)\quad y = −\frac{5}{2} −\frac{x}{10},\: (25, −5)$$

18. $$y=2x^{2}$$

19. $$y=\frac{cx}{\sqrt{|x^{2}=1|}}$$

20. $$y=y_{1}+c(x-x_{1})$$

21. $$y=-\frac{x^{3}}{2}-\frac{x}{2}$$

22. $$y=-x\ln |x|+cx$$

23. $$y=\sqrt{2x+4}$$

24. $$y=\sqrt{x^{2}-3}$$

25. $$y=kx^{2}$$

26. $$(y-x)^{3}(y+x)=k$$

27. $$y^{2}=-x+k$$

28. $$y^{2}=-\frac{1}{2}\ln (1+2x^{2})+k$$

29. $$y^{2}=-2x-\ln (x-1)^{2}+k$$

30. $$y=1+\sqrt{\frac{9-x^{2}}{2}};\text{ those with }c>0$$

33. $$\tan ^{-1}\frac{y}{x}-\frac{1}{2}\ln (x^{2}+y^{2})=k$$

34. $$\frac{1}{2}\ln (x^{2}+y^{2})+(\tan\alpha )\tan ^{-1}\frac{y}{x}=k$$

This page titled 11.31: A.4.5- Section 4.5 Answers is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.