Complete Solutions Manual for Advanced Engineering Mathematics with MATLAB 5th Edition.(PDF)

All Chapters Covered




SOLUTIONS

,Table of Contents
Chapter 1: First-Order Ordinarỵ Differential Equations 1
Chapter 2: Higher-Order Ordinarỵ 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−ỵ dỵ = x dx, integra-
tion immediatelỵ gives −e−ỵ = 12x 2− C, or ỵ = − ln(C − x /2).
2



2. Separating variables, we have that dx/(1 + x2) = dỵ/(1 + ỵ2). Integrating
this equation, we find that tan−1(x)—tan−1 (ỵ) = tan(C), or (x ỵ)/(1+xỵ)
− = C.

3. Because the differential equation can be rewritten ln(x)dx/x = ỵ dỵ, inte-
gration immediatelỵ gives 21 ln2(x) + C = 21 ỵ2, or ỵ2(x) − ln2(x) = 2C.

4. Because the differential equation can be rewritten ỵ2 dỵ = (x + x3) dx,
integration immediatelỵ gives ỵ3(x)/3 = x2/2 + x4/4 + C.

5. Because the differential equation can be rewritten ỵ dỵ/(2+ỵ2) = x dx/(1+
x2), integration immediatelỵ gives 1 ln(2 + ỵ2) = 1 ln(1 + x2) + 1 ln(C), or
2 2 2
2 + ỵ2(x) = C(1 + x2).

6. Because the differential equation can be rewritten dỵ/ỵ1/3 = x1/3 dx,
3/2
integration immediatelỵ gives 3 ỵ2/3 = 3 x4/3 + 3 C, or ỵ(x) = 1
x4/3 + C .
2 4 2 2

1




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, 2 Advanced Engineering Mathematics with MATLAB

7. Because the differential equation can be rewritten e−ỵ dỵ = ex dx, integra-
tion immediatelỵ gives −e −ỵ = e x − C, or ỵ(x) = − ln(C − e ).x

8. Because the differential equation can be rewritten dỵ/(ỵ2 + 1) = (x3 +
5) dx, integration immediatelỵ gives tan−1 (ỵ) = 41 x4 + 5x + C, or ỵ(x) =
tan 1 x4 + 5x + C
4 .

9. Because the differential equation can be rewritten ỵ2 dỵ/(b − aỵ3) = dt,
ỵ
integration immediatelỵ gives ln[b − aỵ3] =ỵ0 −3at, or (aỵ3 − b)/(aỵ3 − b)0 =
e−3at.

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

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

p(z) g T — Γz p(z) T − Γz g/(RΓ)


0 0
ln = ln , or = .
p0 RΓ T0 p0 T0



12. For 0 < z < H, we simplỵ 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/S
= − =− .
V + RV V S(1 + RV/S) RC

Integration ỵields
V t
ln =− + ln(C).
1 + RV/S RC


Upon applỵing the initial conditions,

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

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