2.1E: Linear First Order Equations (Exercises)
( \newcommand{\kernel}{\mathrm{null}\,}\)
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 {−2≤x≤2,−2≤y≤2}.
12. y′+3y=1
13. y′+(1x−1)y=−2x
14. y′+2xy=xe−x2
15. y′+2x1+x2y=e−x1+x2
Q2.1.4
In Exercises 2.1.16-2.1.24 find the general solution.
16. y′+1xy=7x2+3
17. y′+4x−1y=1(x−1)5+sinx(x−1)4
18. xy′+(1+2x2)y=x3e−x2
19. xy′+2y=2x2+1
20. y′+(tanx)y=cosx
21. (1+x)y′+2y=sinx1+x
22. (x−2)(x−1)y′−(4x−3)y=(x−2)3
23. y′+(2sinxcosx)y=e−sin2x
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. (x−1)y′+3y=1(x−1)3+sinx(x−1)2,y(0)=1
31. xy′+2y=8x2,y(1)=3
32. xy′−2y=−x2,y(1)=1
33. y′+2xy=x,y(0)=3
34. (x−1)y′+3y=1+(x−1)sec2x(x−1)3,y(0)=−1
35. (x+2)y′+4y=1+2x2x(x+2)3,y(−1)=2
36. (x2−1)y′−2xy=x(x2−1),y(0)=4
37. (x2−5)y′−2xy=−2x(x2−5),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=e−xtanxx,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 limt→∞G(t).
44.
(a) Plot a direction field and some integral curves for xy′−2y=−1 on the rectangular region {−1≤x≤1,−.5≤y≤1.5}. 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,x≥0,12+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 xy′−2y=−1,y(0)=12.
(e) Use (b) to show that if x0≠0 and y0 is arbitrary, then the initial value problem xy′−2y=−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 limx→∞f(x)=L. Show that if y is the solution of (A), then limx→∞y(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 y1−y2 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)y′−3tany=−1
(b) ey2(2yy′+2x)=1x2
(c) xy′y+2lny=4x2
(d) y′(1+y)2−1x(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(m−1) are continuous on (a,b), where m is a positive integer, and define
f0=f,fj=f′j−1+pfj−1,1≤j≤m.
Show that
u(j+1)=fjy1,0≤j≤m.