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. 7xy′−2y=−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. y′−xy=x3y3;{−3≤x≤3,−2≤y≤2}
6. y′−1+x3xy=y4;{−2≤x≤2,−2≤y≤2}
Q2.4.3
In Exercises 2.4.7-2.4.11 solve the initial value problem.
7. y′−2y=xy3,y(0)=2√2
8. y′−xy=xy3/2,y(1)=4
9. xy′+y=x4y4,y(1)=1/2
10. y′−2y=2y1/2,y(0)=1
11. y′−4y=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. y′−y=xy1/2,y(0)=4
Q2.4.5
14. You may have noticed that the logistic equation P′=aP(1−αP) from Verhulst’s model for population growth can be written in Bernoulli form as P′−aP=−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 P′−aP=−aα(t)P2, a non-separable Bernoulli equation.
- Assuming that P(0)=P0>0, find P for t>0.
- 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.
- Assuming that limt→∞e−at∫t0α(τ)eaτdτ=L exists (finite or infinite), find limt→∞P(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;{−8≤x≤8,−8≤y≤8}
20. xyy′=x2+2y2;{−4≤x≤4,−4≤y≤4}
21. y′=2y2+x2e−(y/x)22xy;{−8≤x≤8,−8≤y≤8}
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′=y2−3xy−5x2x2,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+yx−y
29. (y′x−y)(ln|y|−ln|x|)=x
30. y′=y3+2xy2+x2y+x3x(y+x)2
31. y′=x+2y2x+y
32. y′=yy−2x
33. y′=xy2+2y3x3+x2y+xy2
34. y′=x3+x2y+3y3x3+3xy2
Q2.4.10
35.
- Find a solution of the initial value problem x2y′=y2+xy−4x2,y(−1)=0 on the interval (−∞,0). Verify that this solution is actually valid on (−∞,∞).
- Use Theorem 2.3.1 to show that (A) has a unique solution on (−∞,0).
- Plot a direction field for the differential equation in (A) on a square {−r≤x≤r,−r≤y≤r}, where r is any positive number. Graph the solution you obtained in (a) on this field.
- Graph other solutions of (A) that are defined on (−∞,∞).
- Graph other solutions of (A) that are defined only on intervals of the form (−∞,a), where is a finite positive number.
36.
- Solve the equation xyy′=x2−xy+y2 implicitly.
- Plot a direction field for (A) on a square {0≤x≤r,0≤y≤r} where r is any positive number.
- Let K be a positive integer. (You may have to try several choices for K.) Graph solutions of the initial value problems xyy′=x2−xy+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′=x2−xy+y2,y(x0)=y0, has a unique solution (i) on (0,∞) or (ii) only on an interval (a,∞), where a>0?
- What can you say about the graph of the solution of (B) as x→∞? (Again, assume that x0>0 and y0>0.)
37.
- Solve the equation y′=2y2−xy+2x2xy+2x2 implicitly.
- Plot a direction field for (A) on a square {−r≤x≤r,−r≤y≤r} 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 {−r≤x≤r, −r≤y≤r}, where r>0. What happens to the direction field as you vary r? Why?
40. Prove: If ad−bc≠0, the equation y′=ax+by+αcx+dy+β can be transformed into the homogeneous nonlinear equation dYdX=aX+bYcX+dY by the substitution x=X−X0, y=Y−Y0, 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+y−32x−y−1
42. y′=2x+y+1x+2y−4
43. y′=−x+3y−14x+y−2
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+x2y−x4)
47. y′=y2e−x+4y+2ex
48. y′=y2+ytanx+tan2xsin2x
49. x(lnx)2y′=−4(lnx)2+ylnx+y2
50. 2x(y+2√x)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 R≡−1, (A) is a Riccati equation.) Let y1 be a known solution and y an arbitrary solution of (A). Let z=y−y1. 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+(1−2ex)y+y2; y1=ex
58. xy′=2−x+(2x−2)y−xy2; y1=1
59. xy′=x3+(1−2x2)y+xy2; y1=x