solutions manual to Advanced Engineering Mathematics with MATLAB, 5th Edition. Duffy (1)

All Chapters Covered




SOLUTIONS

,Table of Contents
Chapter 1: First-Order Ordinary Differential Equations 1
Chapter 2: Higher-Order Ordinary Differential Equations
Chapter 3: Linear Algebra
Chapter 4: Vector Calculus
Chapter 5: Fourier Series
Chapter 6: The Fourier Transform
Chapter 7: The Laplace Transform
Chapter 8: The Wave Equation
Chapter 9: The Heat Equation
Chapter 10: Laplace’s Equation
Chapter 11: The Sturm-Liouville Problem
Chapter 12: Special Functions
Appendix A: Derivation of the Laplacian in Polar Coordinates
Appendix B: Derivation of the Laplacian in Spherical Polar Coordinates

, Solution Manual
Section 1.1

1. first-order, linear 2. first-order, nonlinear
3. first-order, nonlinear 4. third-order, linear
5. second-order, linear 6. first-order, nonlinear
7. third-order, nonlinear 8. second-order, linear
9. second-order, nonlinear 10. first-order, nonlinear
11. first-order, nonlinear 12. second-order, nonlinear
13. first-order, nonlinear 14. third-order, linear
15. second-order, nonlinear 16. third-order, nonlinear

Section 1.2

1. Because the differential equation can be rewritten e−y dy = xdx, integra-
tion immediately gives —e−y = 1 x 2 — C, or y = — ln(C — x 2/2).
2

2. Separating variables, we have that dx/(1 + x2) = dy/(1 + y2).
Integrating this equation, we find—that tan−1(x) tan− 1(y) =—tan(C), or (x
y)/(1+xy) = C.

3. Because the differential equation can be rewritten ln(x)dx/x = y dy, inte-
gration immediately gives 1 ln2(x) + C = 1 y2, or y2(x) — ln2(x) = 2C.
2 2

4. Because the differential equation can be rewritten y2 dy = (x + x3)
dx, integration immediately gives y3(x)/3 = x2/2 + x4/4 + C.

5. Because the differential equation can be rewritten y dy/(2+y2) = xdx/(1+
x2), integration immediately gives 1 ln(2 + y2) = 1 ln(1 + x2) + 1 ln(C), or
2 2 2
2 + y2(x) = C(1 + x2).

6. Because the differential equation can be rewritten dy/y1/3 = x1/3 dx,
integration immediately gives 3 y2/3 = 3 x 4/3 + 3 C, or y(x) = 1 x 4/3 + C
3/2 2 4 2 2
.

1

, 2 Advanced Engineering Mathematics with MATLAB

7. Because the differential equation can be rewritten e−y dy = ex dx, integra-
tion immediately gives —e−y = ex — C, or y(x) = — ln(C — ex).
8. Because the differential equation can be rewritten dy/(y2 + 1) = (x3
+ 5) dx, integration immediately gives tan−1(y) = 1 x4 + 5x + C, or
y(x) =
4
tan 14 x4 + 5x + C .

9. Because the differential equation can be rewritten y2 dy/(b — ay3) = dt,
integration immediately gives ln[b — ay 3] yy0 = —3at, or (ay 3 — b)/(ay03 — b) =
e−3at.

10. Because the differential equation can be written du/u = dx/x2, integra-
tion immediately gives u = Ce−1/x or y(x) = x + Ce−1/x.

— g dz/(RT ).
11. From the hydrostatic equation and ideal gas law, dp/p =
Substituting for T (z),
dp g
=— dz.
p R(T 0 — Γz)
Integrating from 0 to z,

p(z) g T0 — Γz p(z) T0 — Γz g/(RΓ)
ln = ln , or = .
p0 RΓ T0 p0 T0


12. For 0 < z < H, we simply use the previous problem. At z = H,
the ṗressure is
T0 — ΓH g/(RΓ)
ṗ(H) = ṗ0 .
T0
Then we follow the examṗle in the text for an isothermal atmosṗhere for
z ≥ H.

13. Seṗarating variables, we find that
dV dV R dV dt
2/S
= — =— .
V + RV V S(1 + RV/S) RC

Integration yields

V t
ln =— + ln(C).
1 + RV/S RC
Uṗon aṗṗlying the initial conditions,

V0 RV0/S
V (t) = e−t/(RC) + e−t/(RC) V (t).
1 + RV0/S 1 + RV0/S