Solution Manual for Advanced Engineering Mathematics with MATLAB (5th Edition) by Dean G. Duffy – A complete step-by-step solutions companion for the main textbook covering key engineering mathematics topics (ordinary/partial differential equations, linea

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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




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, 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 — x2/2).
2

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

5.2 Because the differential equation 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. Because the differential equation can be rewritten dy/y1/3 = x1/3 dx,
3/2
integration immediately gives
2
y =4 3 x4/3 +2 3 C, or y(x) = 21 x4/3 + C .
3 2/3



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.

11. From the hydrostatic equation and ideal gas law, dp/p = — g dz/(RT ).
Substituting 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)
ln = ln Γz , or = .
p0 RΓ p0 T0
T0


12. For 0 < z < H, we simply use the previous problem. 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 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
Upon applying the initial conditions,

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




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