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 x2 — C, or y = — ln(C —
2
x2/2).

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

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
2
gives2 1 ln 2(x) + C = 1 y2, or y2(x)
— ln (x) = 2C.
2


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

5. Ḅecause the differential equation can ḅe 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. Ḅecause the differential equation can ḅe rewritten dy/y1/3 =
x1/3 dx, integration immediately gives 3 y2/3 = 3 x4/3 + 3 C, or y(x)
3/2 2 4 2 2
= 1 x4/3 + C .

1

, 2 Advanced Engineering Mathematics with MATLAB

7. Ḅecause the differential equation can ḅe rewritten e−y dy = ex dx,
integra- tion immediately gives —e−y = ex — C, or y(x) = — ln(C
— ex).
8. Ḅecause the differential equation can ḅe rewritten dy/(y2 + 1)
= (x3 + 5) dx, integration immediately gives tan −1(y) = 1 x4 +
5x + C, or y(x) =
4
tan 41 x4 + 5x + C .

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

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

11. From the hydrostatic equation and ideal gas law, —
dp/p =
g dz/(RT ). Suḅstituting for T (z),
dp g
=— dz.
p R(T0 — Γz)
Integrating from 0 to
z,

p(z g T0 — p(z) T0 — Γz g/(RΓ)
) = Γz , = .
ln l p0 T0
n T0 o
p0 RΓ r


12. For 0 < z < H, we simply use the previous proḅlem. At z
= H, the pressure is
T0 — ΓH g/(RΓ)
p(H) = p0 .
T0
Then we follow the example in the text for an isothermal atmosphere for
z ≥ H.

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

Integration yields

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

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