Solutions Manual for Advanced Engineering Mathematics with MATLAB (5th Edition, 2022) by Duffy – Covers All Chapters

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 = x dx, integra-
tion immediately gives −e−y = 21 x2 − C, or y = − ln(C − x2 /2).

2. Separating variables, we have that dx/(1 + x2 ) = dy/(1 + y 2 ). 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 12 ln2 (x) + C = 21 y 2 , or y 2 (x) − ln2 (x) = 2C.

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

5. Because the differential equation can be rewritten y dy/(2+y 2 ) = x dx/(1+
x2 ), integration immediately gives 12 ln(2 + y 2 ) = 12 ln(1 + x2 ) + 21 ln(C), or
2 + y 2 (x) = C(1 + x2 ).

6. Because the differential equation can be rewritten dy/y 1/3 = x1/3 dx,

3/2
integration immediately gives 32 y 2/3 = 34 x4/3 + 32 C, or y(x) = 21 x4/3 + C .

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, 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/(y 2 + 1) = (x3 +
5) dx, integration
immediately gives tan−1 (y) = 14 x4 + 5x + C, or y(x) =
tan 14 x4 + 5x + C .

9. Because the differential equation can be rewritten y 2 dy/(b − ay 3 ) = dt,
y
integration immediately gives ln[b − ay 3 ] y0 = −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,
     g/(RΓ)
p(z) g T0 − Γz p(z) T0 − Γz
ln = ln , or = .
p0 RΓ T0 p0 T0


12. For 0 < z < H, we simply use the previous problem. At z = H, the
pressure is
 g/(RΓ)
T0 − ΓH
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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