Solutions Manual for Advanced Engineering Mathematics with MATLAB (5th Edition) by Dean G. Duffy

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 = 1 x2 − C, or2 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 1 ln2(x) + C 2= 1 y2 , or y2(x) − 2ln2(x) = 2C.

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. Because the differential equation can be rewritten y dy/(2+y2 ) = x dx/(1+ x2),
integration immediately gives 1 ln(2 + y22 ) = 1 ln(1 + x2) 2+ 1 ln(C), or 2
2 + y2(x) = C(1 + x2).

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

1




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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
4
+ C, or y(x) =
tan 14x4 + 5x + C .

9. Because the differential equation can be rewritten y2 dy/(b − ay3 ) = dt,
y
integration immediately gives ln[b − ay3 ] = −3at, or (ay3 − b)/(ay3 − b) = 0
y0
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.
= − R(T0 − Γz)
p
Integrating from 0 to z,
g/(RΓ)
p(z) p(z) T0 − Γz
ln = g ln T0 − Γz , 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)
p(H) = p 0 T0 − ΓH .
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
= − =− .
V + RV 2/S V S(1 + RV/S) RC

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

Upon applying the initial conditions,

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




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,Worked Solutions 3

Solving for V (t), we obtain

V (t) = SV0 e−t/(RC) .
S + RV0 1 − e −t/(RC)


14. From the definition of γ, we can write the differential equation
A dT
+ T4 = γ4 ,
B dt
or
B dT T 4 1 dT
dT T 2
= − 2 2
dt = − − γ4 + γ2
2γ2 T −γ
A dT
1 2γ dT T 2 dT
= 3 — + .
4γ + γ 2 T −γ T +γ
The final answer follows from direction integration.

15. Separating the variables yields
dN d[ln(K/N )]
= bdt, or = −b dt.
N ln(K/N ) ln(K/N )

Integration leads to

ln [ln(K/N )] − ln {ln[K/N (0)]} = −bt

or
ln {ln(K/N)/ln[K/N(0)]}= −bt ln(K/N) =

or
ln[K/N(0)]e−bt ln[N/N(0)] = ln[K/N(0)]

or 1 − e−bt

or
N (t) = N (0) exp ln[K/N (0)] 1 − e −bt .

16. Separating the variables yields
dI β dI
− = −α dz.
I α 1 + βI/α

Integration leads to

I(z) 1 + βI(0)/α
ln 1 + βI(z)/α = −αz,
I(0)




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or
I(z) I(0) αI(0)e−αz
= e −αz , or I(z) = .
1 + βI(z)/α 1 + βI(0)/α α + βI(0) [1 − e−αz]


17. Separating the variables yields
d[X]
= k dt
([A]0 − [X]) ([B]0 − [X]) ([C]0 − [X])
d[X]
([A]0 − [B]0) ([A]0 − [C]0) ([A]0 − [X])
d[X]
+
([B]0 − [A]0) ([B]0 − [C]0) ([B]0 − [X])
d[X]
+ = k dt
([C]0 − [A]0) ([C]0 − [B]0) ([C]0 − [X])
Integration yields

1 [A]0 [A]0
([A]0 − [B]0) ([A]0 − [C]0) ln
− [X]
1 [B]0
+ ln
([B]0 − [A]0) ([B]0 − [C]0) [B]0 − [X]
1 [C]0 [C]0
+ ln = kt.
([C]0 − [A]0) ([C]0 − [B]0) − [X]


18. Separation of variables yields
d[X]
= (k1 + k2) dt.
α − [X]

Integrating both sides,

ln(α − [X]) − ln(α − [X]0) = −(k1 + k2)t.

Because [X]0 = 0,
h i
α − [X] = αe −(k 1+k 2)t, or [X] = α 1 −e
−(k 1+k 2)t
.


Section 1.3

1. Because M (x, y) = −y and N (x, y) = x + y, we have that M (tx, ty) =
−ty = tM (x, y), and N (tx, ty) = tx + ty = tN (x, y). Therefore, the differen- tial equation is
homogeneous.




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,Worked Solutions 5

Let y = ux. Substituting into the differential equation, (ux + x)(u dx +
x du) = ux dx, or −u2x dx = (1 + u)x2 du, or
dx 1 1
+ du.
— = u u2
x
Integrating this last equation,
1 x
— ln |x| = ln(u) − − C, or ln |y| − = C.
u y

2. Because M (x, y) = y − x and N (x, y) = x + y, we have that M (tx, ty) = ty − tx = tM
(x, y), and N (tx, ty) = tx + ty = tN (x, y). Therefore, the differential equation is
homogeneous.
Let y = ux. Substituting into the differential equation, (u − 1)x dx +(u + 1)x(u dx + x du)
= 0, or
dx u+1
(u2 + 2u − 1) dx = −(u + 1)x du, or − = du.
x u2 + 2u − 1
Integrating this last equation,
2 y
− ln |x| = 1 ln |u2+2u−1|+C, or x2 +y2 − 1 = y2 +2xy−x2 = C.
2
x2 x

3. Because M (x, y) = x2 + y2 and N (x, y) = 2xy, we have that M (tx, ty) = t2 x2 + t2 y2 =
t2(x2 + y2 ) = t2 M (x, y), and N (tx, ty) = 2t2 xy = t2N (x, y). Therefore, the differential
equation is homogeneous.
Let y = ux. Substituting into the differential equation, 2x(ux)(u dx

+ x du) + (x2 + x2 u 2) dx = 0
or
dx 2u
2xu du + (1 + 3u2) dx = 0, or =− du.
x 1 + 3u2
Integrating this last equation,

ln |x| = −31 ln(1 + 3u2) + ln(C1).

Inverting the logarithms,

|x|(1 + 3y2/x2)1/3 = C1 , or |x|(x2 + 3y2) = C.

4. Because M (x, y) = y(y − x) and N (x, y) = x(x + y), we have that M (tx, ty) =
ty(ty − tx) = t2 M (x, y), and N (tx, ty) = tx(tx + ty) = t2 N (x, y). Therefore, the differential
equation is homogeneous.




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,6 Advanced Engineering Mathematics with MATLAB

Let y = ux. Substituting into the differential equation,

x2u(u − 1) dx + x2(u + 1)(u dx + x du) = 0

or
dx u+1
2u2 dx + (u + 1)x du = 0, or 2 =− du.
x u2
Integrating this last equation,
1 x x
ln |x|2 = − ln |u| + + C, or ln |ux2| = C − , or ln |xy| = C − .
u2 y y
√
5. Because M(x, y) = y + 2 xy and N(x, y) = −x, we have that M(tx, ty) =
p √
ty + 2 t2 xy = ty + 2t xy = tM (x, y), and N (tx, ty) = −tx = tN (x, y).
Therefore, the differential equation is homogeneous.
Let y = ux. Substituting into the differential equation,
√ du dx
x(u dx + x du) = (xu + 2x u ) dx, or √ = .
2 u x

Integrating this last equation,

u1/2 = ln |x| + C, or y = x (ln |x| + C)2 .
p
6. Because M(x, y) = x2 + y 2 − y and N(x, y) = x, we have that M (tx, ty)
p p
= t2 x2 + t2 y2 − ty = t x2 + y2 − y = tM (x, y), and N (tx, ty) = tx =
tN (x, y). Therefore, the differential equation is homogeneous. Let y =
ux. Substituting into the differential equation,
p
x2 + x2 u2 − ux dx + x(x du + u dx) = 0,
or p dx du
x 1 + u2 dx + x2 du = 0, or = −√ .
x 1 + u2
Integrating this last equation,
p
− ln(x) = − ln u + 1 + u2 − ln(C).

Inverting the logarithms,
p p
ux + u2 x2 + x2 = C, or y + x2 + y2 = C.

7. Because M(x, y) = sec(y/x) + y/x and N(x, y) = −1, we have that
M (tx, ty) = sec[(ty)/(tx)] + (ty)/(tx) = sec(y/x) + y/x = M (x, y), and




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,Worked Solutions 7

N(tx, ty) = −1 = N(x, y). Therefore, the differential equation is homoge- neous.
Let y = ux. Substituting into the differential equation,

dx
u dx + x du = [sec(u) + u] dx, or cos(u) du = .
x
Integrating and substituting for u, the final answer is

sin(y/x) − ln |x| = C.

8. Because M (x, y) = e y/x + y/x and N (x, y) = −1, we have that M (tx, ty) = e (ty)/(tx) +
(ty)/(tx) = e y/x + y/x = M (x, y), and N (tx, ty) = −1 = N (x, y). Therefore, the differential
equation is homogeneous.
Let y = ux. Substituting into the differential equation,
dx
u dx + x du = (eu + u) dx, or e−u du = .
x
Integrating and substituting for u, the final answer is

y(x) = −x ln (C − ln |x|) .


Section 1.4

1. Since M (x, y) = y2 − x2 , and N (x, y) = 2xy,
∂M ∂N
= 2y = .
∂y ∂x

The exactness criteria is satisfied.
Now, since
∂u
= y2 − x2 ,
∂x
then u(x, y) = xy2 − 1 x3 + 3f (y). To find f(y), we use

∂u
= 2xy + f′(y) = 2xy.
∂y

Therefore, f′(y) = 0, and u(x, y) = xy2 − 1 x3 = C. 3

2. Since M (x, y) = y − x, and N (x, y) = x + y,
∂M ∂N
=1= .
∂y ∂x




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