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2.4E: Transformation of Nonlinear Equations into Separable Equations (Exercises)

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Q2.4.1

In Exercises 2.4.1-2.4.4 solve the given Bernoulli equation.

1. y+y=y2

2. 7xy2y=x2y6

3. x2y+2y=2e1/xy1/2

4. (1+x2)y+2xy=1(1+x2)y

Q2.4.2

In Exercises 2.4.5 and 2.4.6 find all solutions. Also, plot a direction field and some integral curves on the indicated rectangular region.

5. yxy=x3y3;{3x3,2y2}

6. y1+x3xy=y4;{2x2,2y2}

Q2.4.3

In Exercises 2.4.7-2.4.11 solve the initial value problem.

7. y2y=xy3,y(0)=22

8. yxy=xy3/2,y(1)=4

9. xy+y=x4y4,y(1)=1/2

10. y2y=2y1/2,y(0)=1

11. y4y=48xy2,y(0)=1

Q2.4.4

In Exercises 2.4.12 and 2.4.13 solve the initial value problem and graph the solution.

12. x2y+2xy=y3,y(1)=1/2

13. yy=xy1/2,y(0)=4

Q2.4.5

14. You may have noticed that the logistic equation P=aP(1αP) from Verhulst’s model for population growth can be written in Bernoulli form as PaP=aαP2. This isn’t particularly interesting, since the logistic equation is separable, and therefore solvable by the method studied in Section 2.2. So let’s consider a more complicated model, where a is a positive constant and α is a positive continuous function of t on [0,). The equation for this model is PaP=aα(t)P2, a non-separable Bernoulli equation.

  1. Assuming that P(0)=P0>0, find P for t>0.
  2. Verify that your result reduces to the known results for the Malthusian model where α=0, and the Verhulst model where α is a nonzero constant.
  3. Assuming that limteatt0α(τ)eaτdτ=L exists (finite or infinite), find limtP(t).

Q2.4.6

In Exercises 2.4.15-2.4.18 solve the equation explicitly.

15. y=y+xx

16. y=y2+2xyx2

17. xy3y=y4+x4

18. y=yx+secyx

Q2.4.7

In Exercises 2.4.19-2.4.21 solve the equation explicitly. Also, plot a direction field and some integral curves on the indicated rectangular region.

19. x2y=xy+x2+y2;{8x8,8y8}

20. xyy=x2+2y2;{4x4,4y4}

21. y=2y2+x2e(y/x)22xy;{8x8,8y8}

Q2.4.8

In Exercises 2.4.22-2.4.27 solve the initial value problem.

22. y=xy+y2x2,y(1)=2

23. y=x3+y3xy2,y(1)=3

24. xyy+x2+y2=0,y(1)=2

25. y=y23xy5x2x2,y(1)=1

26. x2y=2x2+y2+4xy,y(1)=1

27. xyy=3x2+4y2,y(1)=3

Q2.4.9

In Exercises 2.4.28-2.4.34 solve the given homogeneous equation implicitly.

28. y=x+yxy

29. (yxy)(ln|y|ln|x|)=x

30. y=y3+2xy2+x2y+x3x(y+x)2

31. y=x+2y2x+y

32. y=yy2x

33. y=xy2+2y3x3+x2y+xy2

34. y=x3+x2y+3y3x3+3xy2

Q2.4.10

35.

  1. Find a solution of the initial value problem x2y=y2+xy4x2,y(1)=0 on the interval (,0). Verify that this solution is actually valid on (,).
  2. Use Theorem 2.3.1 to show that (A) has a unique solution on (,0).
  3. Plot a direction field for the differential equation in (A) on a square {rxr,ryr}, where r is any positive number. Graph the solution you obtained in (a) on this field.
  4. Graph other solutions of (A) that are defined on (,).
  5. Graph other solutions of (A) that are defined only on intervals of the form (,a), where is a finite positive number.

36.

  1. Solve the equation xyy=x2xy+y2 implicitly.
  2. Plot a direction field for (A) on a square {0xr,0yr} where r is any positive number.
  3. Let K be a positive integer. (You may have to try several choices for K.) Graph solutions of the initial value problems xyy=x2xy+y2,y(r/2)=krK, for k=1, 2, …, K. Based on your observations, find conditions on the positive numbers x0 and y0 such that the initial value problem xyy=x2xy+y2,y(x0)=y0, has a unique solution (i) on (0,) or (ii) only on an interval (a,), where a>0?
  4. What can you say about the graph of the solution of (B) as x? (Again, assume that x0>0 and y0>0.)

37.

  1. Solve the equation y=2y2xy+2x2xy+2x2 implicitly.
  2. Plot a direction field for (A) on a square {rxr,ryr} where r is any positive number. By graphing solutions of (A), determine necessary and sufficient conditions on (x0,y0) such that (A) has a solution on (i) (,0) or (ii) (0,) such that y(x0)=y0.

38. Follow the instructions of Exercise 2.4.37 for the equation y=xy+x2+y2xy.

39. Pick any nonlinear homogeneous equation y=q(y/x) you like, and plot direction fields on the square {rxr, ryr}, where r>0. What happens to the direction field as you vary r? Why?

40. Prove: If adbc0, the equation y=ax+by+αcx+dy+β can be transformed into the homogeneous nonlinear equation dYdX=aX+bYcX+dY by the substitution x=XX0, y=YY0, where X0 and Y0 are suitably chosen constants.

Q2.4.11

In Exercises 2.4.21-2.4.43 use a method suggested by Exercise 2.4.40 to solve the given equation implicitly.

41. y=6x+y32xy1

42. y=2x+y+1x+2y4

43. y=x+3y14x+y2

Q2.4.12

In Exercises 2.4.44-2.4.51 find a function y1 such that the substitution y=uy1 transforms the given equation into a separable equation of the form (2.4.6). Then solve the given equation explicitly.

44. 3xy2y=y3+x

45. xyy=3x6+6y2

46. x3y=2(y2+x2yx4)

47. y=y2ex+4y+2ex

48. y=y2+ytanx+tan2xsin2x

49. x(lnx)2y=4(lnx)2+ylnx+y2

50. 2x(y+2x)y=(y+x)2

51. (y+ex2)y=2x(y2+yex2+e2x2

Q2.4.13

52. Solve the initial value problem y+2xy=3x2y2+6xy+2x2(2xy+3),y(2)=2.

53. Solve the initial value problem y+3xy=3x4y2+10x2y+6x3(2x2y+5),y(1)=1.

54. Prove: If y is a solution of a homogeneous nonlinear equation y=q(y/x), so is y1=y(ax)/a, where a is any nonzero constant.

55. A generalized Riccati equation is of the form y=P(x)+Q(x)y+R(x)y2. (If R1, (A) is a Riccati equation.) Let y1 be a known solution and y an arbitrary solution of (A). Let z=yy1. Show that z is a solution of a Bernoulli equation with n=2.

Q2.4.14

In Exercises 2.4.56-2.4.59, given that y1 is a solution of the given equation, use the method suggested by Exercise 2.4.55 to find other solutions.

56. y=1+x(1+2x)y+xy2; y1=1

57. y=e2x+(12ex)y+y2; y1=ex

58. xy=2x+(2x2)yxy2; y1=1

59. xy=x3+(12x2)y+xy2; y1=x


This page titled 2.4E: Transformation of Nonlinear Equations into Separable Equations (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.

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