Student Solutions Manual to Accompany Advanced Engineering Mathematics

965 Pages·2006·29.71 MB·English

by  Dennis G. Zill

Table of Contents
Part I Ordinary Differential Equations
1 Introduction to Differential Equations 1
2 First-Order Differential Equations 22
3 Higher-Order Differential Equations 99
4 The Laplace Transform 198
5 Series Solutions of Linear Differential Equations 252
6 Numerical Solutions of Ordinary Differential Equations 317
Part II Vectors, Matrices, and Vector Calculus
7 Vectors 339
8 Matrices 373
9 Vector Calculus 438
Part III Systems of Differential Equations
10 Systems of Linear Differential Equations 551
11 Systems of Nonlinear Differential Equations 604
Part IV Fourier Series and Partial Differential Equations
12 Orthogonal Functions and Fourier Series 634
13 Boundary-Value Problems in Rectangular Coordinates 680
14 Boundary-Value Problems in Other Coordinate Systems 755
15 Integral Transform Method 793
16 Numerical Solutions of Partial Differential Equations 832
Part V Complex Analysis
17 Functions of a Complex Variable 854
18 Integration in the Complex Plane 877
19 Series and Residues 896
20 Conformal Mappings 919
Appendices
Appendix II Gamma function 942
Projects
3.7 Road Mirages 944
3.10 The Ballistic Pendulum 946
8.1 Two-Ports in Electrical Circuits 947
8.2 Traffic Flow 948
8.15 Temperature Dependence of Resistivity 949
9.16 Minimal Surfaces 950
14.3 The Hydrogen Atom 952
15.4 The Uncertainity Inequality in Signal Processing 955
15.4 Fraunhofer Diffraction by a Circular Aperture 958
16.2 Instabilities of Numerical Methods 960
Part I Ordinary Differential Equations
11 Introduction to
Differential Equations
EXERCISES 1.1
Definitions and Terminology
1. Second order; linear
2. Third order; nonlinear because of (dy/dx)4
3. Fourth order; linear
4. Second order; nonlinear because of cos(r+u)
(cid:1)
5. Second order; nonlinear because of (dy/dx)2 or 1+(dy/dx)2
6. Second order; nonlinear because of R2
7. Third order; linear
8. Second order; nonlinear because of x˙2
9. Writing the differential equation in the form x(dy/dx)+y2 =1, we see that it is nonlinear in y because of y2.
However, writing it in the form (y2−1)(dx/dy)+x=0, we see that it is linear in x.
10. Writing the differential equation in the form u(dv/du)+(1+u)v =ueu we see that it is linear in v. However,
writing it in the form (v+uv−ueu)(du/dv)+u=0, we see that it is nonlinear in u.
11. From y =e−x/2 we obtain y(cid:1) =−1e−x/2. Then 2y(cid:1)+y =−e−x/2+e−x/2 =0.
2
12. From y = 6 − 6e−20t we obtain dy/dt=24e−20t, so that
5 5 (cid:2) (cid:3)
dy 6 6
+20y =24e−20t+20 − e−20t =24.
dt 5 5
13. From y = e3xcos2x we obtain y(cid:1) = 3e3xcos2x − 2e3xsin2x and y(cid:1)(cid:1) = 5e3xcos2x − 12e3xsin2x, so that
y(cid:1)(cid:1)−6y(cid:1)+13y =0.
14. From y =−cosxln(secx+tanx) we obtain y(cid:1) =−1+sinxln(secx+tanx) and
y(cid:1)(cid:1) =tanx+cosxln(secx+tanx). Then y(cid:1)(cid:1)+y =tanx.
15. The domain of the function, found by solving x+2≥0, is [−2,∞). From y(cid:1) =1+2(x+2)−1/2 we have
(y−x)y(cid:1) =(y−x)[1+(2(x+2)−1/2]
=y−x+2(y−x)(x+2)−1/2
=y−x+2[x+4(x+2)1/2−x](x+2)−1/2
=y−x+8(x+2)1/2(x+2)−1/2 =y−x+8.
1
1.1 Definitions and Terminology
An interval of definition for the solution of the differential equation is (−2,∞) because y(cid:1) is not defined at
x=−2.
16. Since tanx is not defined for x = π/2 + nπ, n an integer, the domain of y = 5tan5x is
(cid:4) (cid:4)
{x(cid:4)5x(cid:3)=π/2+nπ} or {x(cid:4)x(cid:3)=π/10+nπ/5}. From y(cid:1) =25sec25x we have
y(cid:1) =25(1+tan25x)=25+25tan25x=25+y2.
An interval of definition for the solution of the differential equation is (−π/10,π/10). Another interval is
(π/10,3π/10), and so on.
(cid:4) (cid:4)
17. The domain of the function is {x(cid:4)4−x2 (cid:3)=0} or {x(cid:4)x(cid:3)=−2 or x(cid:3)=2}. From y(cid:1) =2x/(4−x2)2 we have
(cid:2) (cid:3)
1 2
(cid:1)
y =2x =2xy.
4−x2
An interval of definition for the solution of the differential equation is (−2,2). Other intervals are (−∞,−2)
and (2,∞).
√
18. The function is y =1/ 1−sinx, whose domain is obtained from 1−sinx(cid:3)=0 or sinx(cid:3)=1. Thus, the domain
(cid:4)
is {x(cid:4)x(cid:3)=π/2+2nπ}. From y(cid:1) =−1(1−sinx)−3/2(−cosx) we have
2
2y(cid:1) =(1−sinx)−3/2cosx=[(1−sinx)−1/2]3cosx=y3cosx.
Anintervalofdefinitionforthesolutionofthedifferentialequationis(π/2,5π/2). Anotheroneis(5π/2,9π/2),
and so on.
19. Writingln(2X−1)−ln(X−1)=tanddifferentiatingimplicitlyweobtain
X
2 dX 1 dX
2X −1 dt − X−1 dt =1 4
(cid:2) (cid:3)
2 1 dX
− =1 2
2X−1 X−1 dt
2X −2−2X+1 dX
t
=1 -4 -2 2 4
(2X −1)(X−1) dt
dX =−(2X −1)(X−1)=(X −1)(1−2X). -2
dt
Exponentiating both sides of the implicit solution we obtain -4
2X−1
=et
X−1
2X−1=Xet−et
(et−1)=(et−2)X
et−1
X = .
et−2
Solving et−2=0 we get t=ln2. Thus, the solution is defined on (−∞,ln2) or on (ln2,∞). The graph of the
solution defined on (−∞,ln2) is dashed, and the graph of the solution defined on (ln2,∞) is solid.
2
1.1 Definitions and Terminology
20. Implicitly differentiating the solution, we obtain
y
dy dy
−2x2 −4xy+2y =0
4
dx dx
−x2dy−2xydx+ydy =0
2
2xydx+(x2−y)dy =0.
Using the quadratic formula to solve y2 −2x2y −1 = 0 for y, we get x
(cid:5) √ (cid:6) √ -4 -2 2 4
y = 2x2 ± 4x4+4 /2 = x2 ± x4+1. Thus, two explicit solutions
√ √
arey =x2+ x4+1 andy =x2− x4+1. Bothsolutionsaredefined -2
1 2
on (−∞,∞). The graph of y (x) is solid and the graph of y is dashed.
1 2
-4
21. Differentiating P =c et/(1+c et) we obtain
1 1
dP (1+c et)c et−c et·c et c et [(1+c et)−c et]
= 1 1 1 1 = 1 1 1
dt (1+c1et)2 1+c1et 1+c1et
(cid:7) (cid:8)
c et c et
= 1 1− 1 =P(1−P).
1+c et 1+c et
1 1
(cid:9)
x
22. Differentiating y =e−x2 et2dt+c e−x2 we obtain
1
0
(cid:9) (cid:9)
x x
y(cid:1) =e−x2ex2 −2xe−x2 et2dt−2c xe−x2 =1−2xe−x2 et2dt−2c xe−x2.
1 1
0 0
Substituting into the differential equation, we have
(cid:9) (cid:9)
x x
y(cid:1)+2xy =1−2xe−x2 et2dt−2c xe−x2 +2xe−x2 et2dt+2c xe−x2 =1.
1 1
0 0
dy d2y
23. From y =c e2x+c xe2x we obtain =(2c +c )e2x+2c xe2x and =(4c +4c )e2x+4c xe2x, so that
1 2 dx 1 2 2 dx2 1 2 2
d2y dy
−4 +4y =(4c +4c −8c −4c +4c )e2x+(4c −8c +4c )xe2x =0.
dx2 dx 1 2 1 2 1 2 2 2
24. From y =c x−1+c x+c xlnx+4x2 we obtain
1 2 3
dy
=−c x−2+c +c +c lnx+8x,
1 2 3 3
dx
d2y
=2c x−3+c x−1+8,
dx2 1 3
and
d3y
=−6c x−4−c x−2,
dx3 1 3
so that
d3y d2y dy
x3 +2x2 −x +y =(−6c +4c +c +c )x−1+(−c +2c −c −c +c )x
dx3 dx2 dx 1 1 1 1 3 3 2 3 2
+(−c +c )xlnx+(16−8+4)x2
3 3
=12x2.
(cid:10) (cid:10)
−x2, x<0 −2x, x<0
25. From y = we obtain y(cid:1) = so that xy(cid:1)−2y =0.
x2, x≥0 2x, x≥0
3
1.1 Definitions and Terminology
26. The function y(x) is not continuous at x = 0 since lim y(x) = 5 and lim y(x) = −5. Thus, y(cid:1)(x) does not
x→0− x→0+
exist at x=0.
27. (a) From y =emx we obtain y(cid:1) =memx. Then y(cid:1)+2y =0 implies
memx+2emx =(m+2)emx =0.
Since emx >0 for all x, m=−2. Thus y =e−2x is a solution.
(b) From y =emx we obtain y(cid:1) =memx and y(cid:1)(cid:1) =m2emx. Then y(cid:1)(cid:1)−5y(cid:1)+6y =0 implies
m2emx−5memx+6emx =(m−2)(m−3)emx =0.
Since emx >0 for all x, m=2 and m=3. Thus y =e2x and y =e3x are solutions.
28. (a) From y =xm we obtain y(cid:1) =mxm−1 and y(cid:1)(cid:1) =m(m−1)xm−2. Then xy(cid:1)(cid:1)+2y(cid:1) =0 implies
xm(m−1)xm−2+2mxm−1 =[m(m−1)+2m]xm−1 =(m2+m)xm−1
=m(m+1)xm−1 =0.
Since xm−1 >0 for x>0, m=0 and m=−1. Thus y =1 and y =x−1 are solutions.
(b) From y =xm we obtain y(cid:1) =mxm−1 and y(cid:1)(cid:1) =m(m−1)xm−2. Then x2y(cid:1)(cid:1)−7xy(cid:1)+15y =0 implies
x2m(m−1)xm−2−7xmxm−1+15xm =[m(m−1)−7m+15]xm
=(m2−8m+15)xm =(m−3)(m−5)xm =0.
Since xm >0 for x>0, m=3 and m=5. Thus y =x3 and y =x5 are solutions.
In Problems 29–32, we substitute y =c into the differential equations and use y(cid:1) =0 and y(cid:1)(cid:1) =0
29. Solving 5c=10 we see that y =2 is a constant solution.
30. Solving c2+2c−3=(c+3)(c−1)=0 we see that y =−3 and y =1 are constant solutions.
31. Since 1/(c−1)=0 has no solutions, the differential equation has no constant solutions.
32. Solving 6c=10 we see that y =5/3 is a constant solution.
33. From x=e−2t+3e6t and y =−e−2t+5e6t we obtain
dx dy
=−2e−2t+18e6t and =2e−2t+30e6t.
dt dt
Then
dx
x+3y =(e−2t+3e6t)+3(−e−2t+5e6t)=−2e−2t+18e6t =
dt
and
dy
5x+3y =5(e−2t+3e6t)+3(−e−2t+5e6t)=2e−2t+30e6t = .
dt
34. From x=cos2t+sin2t+ 1et and y =−cos2t−sin2t− 1et we obtain
5 5
dx 1 dy 1
=−2sin2t+2cos2t+ et and =2sin2t−2cos2t− et
dt 5 dt 5
and
d2x 1 d2y 1
=−4cos2t−4sin2t+ et and =4cos2t+4sin2t− et.
dt2 5 dt2 5
Then
1 1 d2x
4y+et =4(−cos2t−sin2t− et)+et =−4cos2t−4sin2t+ et =
5 5 dt2
and
4
1.1 Definitions and Terminology
1 1 d2y
4x−et =4(cos2t+sin2t+ et)−et =4cos2t+4sin2t− et = .
5 5 dt2
35. (y(cid:1))2+1=0 has no real solutions because (y(cid:1))2+1 is positive for all functions y =φ(x).
36. The only solution of (y(cid:1))2+y2 =0 is y =0, since if y (cid:3)=0, y2 >0 and (y(cid:1))2+y2 ≥y2 >0.
37. The first derivative of f(x)=ex is ex. The first derivative of f(x)=ekx is kekx. The differential equations are
y(cid:1) =y and y(cid:1) =ky, respectively.
38. Any function of the form y = cex or y = ce−x is its own second derivative. The corresponding differential
equation is y(cid:1)(cid:1) −y = 0. Functions of the form y = csinx or y = ccosx have second derivatives that are the
negatives of themselves. The differential equation is y(cid:1)(cid:1)+y =0.
(cid:1) (cid:1) √
39. We first note that 1−y2 = 1−sin2x = cos2x = |cosx|. This prompts us to consider values of x for
which cosx<0, such as x=π. In this case
(cid:4) (cid:4)
dy(cid:4)(cid:4)(cid:4) = d (sinx)(cid:4)(cid:4)(cid:4) =cosx(cid:4)(cid:4) =cosπ =−1,
dx(cid:4) dx (cid:4) x=π
x=π x=π
but (cid:1) (cid:1) √
1−y2| = 1−sin2π = 1=1.
x=π
(cid:1)
Thus, y = sinx will only be a solution of y(cid:1) = 1−y2 when cosx > 0. An interval of definition is then
(−π/2,π/2). Other intervals are (3π/2,5π/2), (7π/2,9π/2), and so on.
40. Since the first and second derivatives of sint and cost involve sint and cost, it is plausible that a linear
combination of these functions, Asint+Bcost, could be a solution of the differential equation. Using y(cid:1) =
Acost−Bsint and y(cid:1)(cid:1) =−Asint−Bcost and substituting into the differential equation we get
y(cid:1)(cid:1)+2y(cid:1)+4y =−Asint−Bcost+2Acost−2Bsint+4Asint+4Bcost
=(3A−2B)sint+(2A+3B)cost=5sint.
Thus 3A−2B = 5 and 2A+3B = 0. Solving these simultaneous equations we find A = 15 and B = −10. A
13 13
particular solution is y = 15sint− 10cost.
13 13
41. Onesolutionisgivenbytheupperportionofthegraphwithdomainapproximately(0,2.6). Theothersolution
is given by the lower portion of the graph, also with domain approximately (0,2.6).
42. Onesolution,withdomainapproximately(−∞,1.6)istheportionofthegraphinthesecondquadranttogether
with the lower part of the graph in the first quadrant. A second solution, with domain approximately (0,1.6)
is the upper part of the graph in the first quadrant. The third solution, with domain (0,∞), is the part of the
graph in the fourth quadrant.
43. Differentiating (x3+y3)/xy =3c we obtain
xy(3x2+3y2y(cid:1))−(x3+y3)(xy(cid:1)+y)
=0
x2y2
3x3y+3xy3y(cid:1)−x4y(cid:1)−x3y−xy3y(cid:1)−y4 =0
(3xy3−x4−xy3)y(cid:1) =−3x3y+x3y+y4
y4−2x3y y(y3−2x3)
(cid:1)
y = = .
2xy3−x4 x(2y3−x3)
44. A tangent line will be vertical where y(cid:1) is undefined, or in this case, where x(2y3−x3) = 0. This gives x = 0
and 2y3 =x3. Substituting y3 =x3/2 into x3+y3 =3xy we get
5
1.1 Definitions and Terminology
(cid:2) (cid:3)
1 1
x3+ x3 =3x x
2 21/3
3 3
x3 = x2
2 21/3
x3 =22/3x2
x2(x−22/3)=0.
Thus, there are vertical tangent lines at x=0 and x=22/3, or at (0,0) and (22/3,21/3). Since 22/3 ≈1.59, the
estimates of the domains in Problem 42 were close.
√ √
45. The derivatives of the functions are φ(cid:1)(x)=−x/ 25−x2 and φ(cid:1)(x)=x/ 25−x2, neither of which is defined
1 2
at x=±5.
46. Todetermineifasolutioncurvepassesthrough(0,3)welett=0andP =3intheequationP =c et/(1+c et).
1 1
This gives 3=c /(1+c ) or c =−3. Thus, the solution curve
1 1 1 2
(−3/2)et −3et
P = =
1−(3/2)et 2−3et
passes through the point (0,3). Similarly, letting t=0 and P =1 in the equation for the one-parameter family
of solutions gives 1 = c /(1+c ) or c = 1+c . Since this equation has no solution, no solution curve passes
1 1 1 1
through (0,1).
47. Forthefirst-orderdifferentialequationintegratef(x). Forthesecond-orderdifferentialequationintegratetwice.
(cid:11) (cid:11)
In the latter case we get y = ( f(x)dx)dx+c x+c .
1 2
48. Solving for y(cid:1) using the quadratic formula we obtain the two differential equations
(cid:12) (cid:1) (cid:13) (cid:12) (cid:1) (cid:13)
1 1
y(cid:1) = 2+2 1+3x6 and y(cid:1) = 2−2 1+3x6 ,
x x
so the differential equation cannot be put in the form dy/dx=f(x,y).
49. The differential equation yy(cid:1)−xy =0 has normal form dy/dx=x. These are not equivalent because y =0 is a
solution of the first differential equation but not a solution of the second.
50. Differentiating we get y(cid:1) =c +3c x2 and y(cid:1)(cid:1) =6c x. Then c =y(cid:1)(cid:1)/6x and c =y(cid:1)−xy(cid:1)(cid:1)/2, so
1 2 2 2 1
(cid:2) (cid:3) (cid:2) (cid:3)
xy(cid:1)(cid:1) y(cid:1)(cid:1) 1
y = y(cid:1)− x+ x3 =xy(cid:1)− x2y(cid:1)(cid:1)
2 6x 3
and the differential equation is x2y(cid:1)(cid:1)−3xy(cid:1)+3y =0.
51. (a) Sincee−x2 ispositiveforallvaluesofx,dy/dx>0forallx,andasolution,y(x),ofthedifferentialequation
must be increasing on any interval.
dy dy
(b) lim = lim e−x2 =0 and lim = lim e−x2 =0. Since dy/dx approaches 0 as x approaches −∞
x→−∞dx x→−∞ x→∞dx x→∞
and ∞, the solution curve has horizontal asymptotes to the left and to the right.
(c) To test concavity we consider the second derivative
(cid:2) (cid:3) (cid:12) (cid:13)
d2y d dy d
= = e−x2 =−2xe−x2.
dx2 dx dx dx
Since the second derivative is positive for x<0 and negative for x>0, the solution curve is concave up on
(−∞,0) and concave down on (0,∞).
6
1.1 Definitions and Terminology
(d) y
x
52. (a) The derivative of a constant solution y = c is 0, so solving 5−c = 0 we see that c = 5 and so y = 5 is a
constant solution.
(b) Asolutionisincreasingwheredy/dx=5−y >0ory <5. Asolutionisdecreasingwheredy/dx=5−y <0
or y >5.
53. (a) The derivative of a constant solution is 0, so solving y(a−by) = 0 we see that y = 0 and y = a/b are
constant solutions.
(b) Asolutionisincreasingwheredy/dx=y(a−by)=by(a/b−y)>0or0<y <a/b. Asolutionisdecreasing
where dy/dx=by(a/b−y)<0 or y <0 or y >a/b.
(c) Using implicit differentiation we compute
d2y
=y(−by(cid:1))+y(cid:1)(a−by)=y(cid:1)(a−2by).
dx2
Solving d2y/dx2 = 0 we obtain y = a/2b. Since d2y/dx2 > 0 for 0 < y < a/2b and d2y/dx2 < 0 for
a/2b<y <a/b, the graph of y =φ(x) has a point of inflection at y =a/2b.
(d)
y
y=a b
ê
y=0
x
54. (a) If y =c is a constant solution then y(cid:1) =0, but c2+4 is never 0 for any real value of c.
(b) Since y(cid:1) =y2+4>0 for all x where a solution y =φ(x) is defined, any solution must be increasing on any
interval on which it is defined. Thus it cannot have any relative extrema.
(c) Using implicit differentiation we compute d2y/dx2 = 2yy(cid:1) = 2y(y2+4). Setting d2y/dx2 = 0 we see that
y = 0 corresponds to the only possible point of inflection. Since d2y/dx2 < 0 for y < 0 and d2y/dx2 > 0
for y >0, there is a point of inflection where y =0.
7