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2.1E: Linear First Order Equations (Exercises)

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Q2.1.1

In Exercises 2.1.1-2.1.5 find the general solution.

1. y+ay=0 ( a=constant)

2. y+3x2y=0

3. xy+(lnx)y=0

4. xy+3y=0

5. x2y+y=0

Q2.1.2

In Exercises 2.1.6-2.1.11 solve the initial value problem.

6. y+(1+xx)y=0,y(1)=1

7. xy+(1+1lnx)y=0,y(e)=1

8. xy+(1+xcotx)y=0,y(π2)=2

9. y(2x1+x2)y=0,y(0)=2

10. y+kxy=0,y(1)=3(k=constant)

11. y+(tankx)y=0,y(0)=2(k=constant)

Q2.1.3

In Exercises 2.1.12-2.1.15 find the general solution. Also, plot a direction field and some integral curves on the rectangular region {2x2,2y2}.

12. y+3y=1

13. y+(1x1)y=2x

14. y+2xy=xex2

15. y+2x1+x2y=ex1+x2

Q2.1.4

In Exercises 2.1.16-2.1.24 find the general solution.

16. y+1xy=7x2+3

17. y+4x1y=1(x1)5+sinx(x1)4

18. xy+(1+2x2)y=x3ex2

19. xy+2y=2x2+1

20. y+(tanx)y=cosx

21. (1+x)y+2y=sinx1+x

22. (x2)(x1)y(4x3)y=(x2)3

23. y+(2sinxcosx)y=esin2x

24. x2y+3xy=ex

Q2.1.5

In Exercises 2.1.25-2.1.29 solve the initial value problem and sketch the graph of the solution.

25. y+7y=e3x,y(0)=0

26. (1+x2)y+4xy=21+x2,y(0)=1

27. xy+3y=2x(1+x2),y(1)=0

28. y+(cotx)y=cosx,y(π2)=1

29. y+1xy=2x2+1,y(1)=0

Q2.1.6

In Exercises 2.1.30-2.1.37, solve the initial value problem.

30. (x1)y+3y=1(x1)3+sinx(x1)2,y(0)=1

31. xy+2y=8x2,y(1)=3

32. xy2y=x2,y(1)=1

33. y+2xy=x,y(0)=3

34. (x1)y+3y=1+(x1)sec2x(x1)3,y(0)=1

35. (x+2)y+4y=1+2x2x(x+2)3,y(1)=2

36. (x21)y2xy=x(x21),y(0)=4

37. (x25)y2xy=2x(x25),y(2)=7

Q2.1.7

In Excercises 2.1.28-2.1.42 solve the initial value problem and leave the answer in a form involving a definite integral. (You can solve these problems numerically by methods discussed in Chapter 3.)

38. y+2xy=x2,y(0)=3

39. y+1xy=sinxx2,y(1)=2

40. y+y=extanxx,y(1)=0

41. y+2x1+x2y=ex(1+x2)2,y(0)=1

42. xy+(x+1)y=ex2,y(1)=2

43. Experiments indicate that glucose is absorbed by the body at a rate proportional to the amount of glucose present in the bloodstream. Let λ denote the (positive) constant of proportionality. Now suppose glucose is injected into a patient’s bloodstream at a constant rate of r units per unit of time. Let G=G(t) be the number of units in the patient’s bloodstream at time t>0. Then G=λG+r, where the first term on the right is due to the absorption of the glucose by the patient’s body and the second term is due to the injection. Determine G for t>0, given that G(0)=G0. Also, find limtG(t).

44.

(a) Plot a direction field and some integral curves for xy2y=1 on the rectangular region {1x1,.5y1.5}. What do all the integral curves have in common?

(b) Show that the general solution of (A) on (,0) and (0,) is

y=12+cx2.

(c) Show that y is a solution of (A) on (,) if and only if y={12+c1x2,x0,12+c2x2,x<0, where c1 and c2 are arbitrary constants.

(d) Conclude from c that all solutions of (A) on (,) are solutions of the initial value problem xy2y=1,y(0)=12.

(e) Use (b) to show that if x00 and y0 is arbitrary, then the initial value problem xy2y=1,y(x0)=y0 has infinitely many solutions on ( ,). Explain why this doesn't contradict Theorem 2.1.1.

45. Suppose f is continuous on an open interval (a,b) and α is a constant.

(a) Derive a formula for the solution of the initial value problem

y+αy=f(x),y(x0)=y0,

where x0 is in (a,b) and y0 is an arbitrary real number.

(b) Suppose (a,b)=(a,), α>0 and limxf(x)=L. Show that if y is the solution of (A), then limxy(x)=L/α.

46. Assume that all functions in this exercise are defined on a common interval (a,b).

(a) Prove: If y1 and y2 are solutions of

y+p(x)y=f1(x)

and

y+p(x)y=f2(x)

respectively, and c1 and c2 are constants, then y=c1y1+c2y2 is a solution of

y+p(x)y=c1f1(x)+c2f2(x).

(This is the principle of superposition.)

(b) Use (a) to show that if y1 and y2 are solutions of the nonhomogeneous equation

y+p(x)y=f(x),(A)

then y1y2 is a solution of the homogeneous equation

y+p(x)y=0.(B)

(c) Use (a) to show that if y1 is a solution of (A) and y2 is a solution of (B), then y1+y2 is a solution of (A).

47. Some nonlinear equations can be transformed into linear equations by changing the dependent variable. Show that if

g(y)y+p(x)g(y)=f(x)

where y is a function of x and g is a function of y, then the new dependent variable z=g(y) satisfies the linear equation

z+p(x)z=f(x).

48. Solve by the method discussed in Exercise 47.

(a) (sec2y)y3tany=1

(b) ey2(2yy+2x)=1x2

(c) xyy+2lny=4x2

(d) y(1+y)21x(1+y)=3x2

49. We’ve shown that if p and f are continuous on (a,b) then every solution of

y+p(x)y=f(x)

on (a,b) can be written as y=uy1, where y1 is a nontrivial solution of the complementary equation for (A) and u=f/y1. Now suppose f, f, …, f(m) and p, p, …, p(m1) are continuous on (a,b), where m is a positive integer, and define

f0=f,fj=fj1+pfj1,1jm.

Show that

u(j+1)=fjy1,0jm.


This page titled 2.1E: Linear First Order Equations (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform.

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