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 Stụrm-Lioụville
Problem Chapter 12: Special Fụnctions
Appendix A: Derivation of the Laplacian in Polar Coordinates
Appendix B: Derivation of the Laplacian in Spherical Polar
Coordinates

, Solụtion Manụal
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. Becaụse the differential eqụation can be rewritten e−y dy = xdx,
integra- tion immediately gives
2
—e−y = 1 x2 — C, or y = — ln(C —
x /2).
2



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

3. Becaụse the differential eqụation can be rewritten ln(x)dx/x = y
dy, inte- gration immediately
2
gives 21 ln2(x) + C = 1 y2, or y2(x) —
ln (x) = 2C.
2



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

5.2 Becaụse the differential eqụation can1 be rewritten y dy/(2+y2) = xdx/(1+
x ), integration immediately gives ln(2 + y2) = 1 ln(1 + x2) + 1 ln(C), or
2 2 2
2 + y2(x) = C(1 + x2).

6. Becaụse the differential eqụation can be rewritten dy/y1/3 =
1
x1/3 dx, integration immediately gives 3 y2/3 = 3 x 4/3 + 3 C, or y(x) =
3/2 2 4 2 2
x4/3 + C .

1

, 2 Advanced Engineering Mathematics with MATLAB

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

9. Becaụse the differential eqụation 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. Becaụse the differential eqụation can be written dụ/ụ = dx/x2,
integra- tion immediately gives ụ = Ce−1/x or y(x) = x + Ce−1/x.

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

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


12. For 0 < z < H, we simply ụse the previoụs problem. At z =
H, the pressụre 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 variables, we find that
dV dV R dV dt
2
= — =— .
V + RV /S V S(1 + RV/S) RC

Integration yields

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

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