Solutions Manual to Advanced Engineering Mathematics with MATLAB 5th Edition by Dean G. Duffy PDF | Comprehensive Step-by-Step Solutions and MATLAB Applications | Covers Differential Equations, Linear Algebra, Fourier and Laplace Transforms, Complex Varia

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




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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 —
2
x2/2).

2. Separating variables, we have that dx/(1 + x2) = dy/(1 + y2).
—
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 immediately2
gives 1 ln2(x)
2
+ C = 1 y2, or y2(x) — ln2(x)
= 2C.

4. Because the differential equation can be rewritten y2 dy = (x +
x3) dx, integration immediately gives y 3(x)/3 = x 2/2 + x4/4 + C.

5. Because the differential equation can be rewritten y dy/(2+y2) = xdx/(1+
x2), integration immediately gives 1 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, integration immediately gives 3 y 2/3 = 3 x4/3 + 3 C, or y(x) = 1 x 4/3
3/2 2 4 2 2
+C .

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 +
C, or y(x) =
4
tan 41 x 4 + 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 = , = .
p0 l Γz p0 T0
n T0 o
RΓ r


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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,Worked Solutions 3

Solving for V (t), we obtain

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

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

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

or
bt ln(K/N ) = ln[K/N (0)]e−bt

or
ln[N/N (0)] = ln[K/N (0)] 1 —
or
e−bt
}
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 = —αz,
1 + βI(z)/α 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
ln
([A]0 — [B]0) ([A]0 — [C]0) [A]0 — [X]
1 [B]0
+ ln
([B]0 — [A]0) ([B]0 — [C]0) [B]0 — [X]
1 = kt.
+ ln
[C]0
([C]0 — [A]0) ([C]0 — [B]0) [C]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−(k1+k2)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 +
xdu) = uxdx, or —u2xdx = (1 + u)x2 du, or

dx 1 1
— = + du.
x u u2
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)xdx
+(u + 1)x(u dx + xdu) = 0, or
dx u+1
(u2 + 2u — 1) dx = —(u + 1)xdu, or — = du.
x u2 + 2u — 1
Integrating this last
equation,

y2 y
— ln |x| = 12 ln |u2 +2u—1|+C, or x
2
2
+ 2 — 1 = y2 +2xy —x 2 = C.
x x

3. Because M (x, y) = x2 + y2 and N (x, y) = 2xy, we have that M (tx,
ty) = t2x2 + t2y2 = t2(x2 + y2) = t2 M (x, y), and N (tx, ty) = 2t2xy = t2N
(x, y). Therefore, the differential equation is homogeneous.
Let y = ux. Substituting into the differential

equation, 2x(ux)(u dx + xdu) + (x2 +

x2u2) dx = 0

or
dx 2u
2xu du + (1 + 3u2) dx = 0, or =— du.
x 1 + 3u2
Integrating this last equation,
1
ln |x| = —
3
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) =
t2N (x, y). Therefore, the differential equation is homogeneous.

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Let y = ux. Substituting into the differential equation,

x2u(u — 1) dx + x2(u + 1)(u dx + xdu) = 0
or
dx u+1
2u2 dx + (u + 1)xdu = 0, or 2 =— du.
x u2
Integrating this last equation,

1 x
ln |x|2 = — ln |u| + + C, or ln |ux2 | = C — , or ln
u2 y y
x
|xy| = C — .

√
5. Because M (x, y) = y + 2 xy and N (x, y) = —x, we have that M (tx,
√
ty) = ty 2+ 2 t 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 + xdu) = (xu + u ) dx, √ = .
2x o 2 u x
r

Integrating this last equation,

u1/2 = ln |x| + C, or y = x (ln |x| + C) 2 .
√
6. Because M (x, y) = x 2 + y 2 — y and N (x, y) = x, we have that M (tx, ty)
√ √
= t2x2 + t2y2 — 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,
√
x2 + x2u2 — ux dx + x(xdu + u dx) = 0,

or
dx du
x√ 1 + u2 dx + x2 du = 0, or = —√ .
x 1 + u2
Integrating this last equation,
√
— ln(x) = — ln u + 1 + u2 — ln(C).

Inverting the logarithms,
√ √
ux + u2x2 + 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 + xdu = [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) = ey/x + y/x and N (x, y)—= 1, we have that M (tx,
ty) = e(ty)/(tx) + (ty)/(tx) = ey/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 + xdu = (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) = xy 2 —3 1 x3 + f (y). To find f (y), we use
∂u
= 2xy + f ′(y) = 2xy.
∂y

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

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

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The exactness criteria is satisfied.
Now, since
∂u
= y — x,
∂x
then u(x, y) = xy — 12x2 + f (y). To find f (y), we use
∂u
= x + f ′(y) = x + y.
∂y
Therefore, f ′(y) = y, and u(x, y) = xy2 + 1 y22 — 1 x2 = C.
3. Since M (x, y) = y2 — 1, and N (x, y) = 2xy — sin(y),
∂M ∂N
= 2y = .
∂y ∂x
The exactness criteria is
satisfied.
Now, since
∂u
= y2 — 1,
∂x
then u(x, y) = xy2 — x + f (y). To find f (y), we use
∂u
= 2xy + f ′(y) = 2xy— sin(y).
∂y
Therefore, f ′(y) = — sin(y), and u(x, y) = xy2 — x + cos(y) = C.
4. Since M (x, y) = sin(y) — 2xy + x2, and N (x, y) = x cos(y) — x2,
∂M ∂N
= cos(y) — 2x = .
∂y ∂x
The exactness criteria is satisfied.
Now, since
∂u
= sin(y)— 2xy + x2,
∂x
then u(x, y) = x sin(y) — x 2y 3+ 1 x 3 + f (y). To find f (y), we use
∂u
= x cos(y)
— x + f (y) = x cos(y)
2 ′
—x .
2
∂y
Therefore, f ′(y) = 0, and u(x, y) = x sin(y) — x 2y + 1 x3 = C.
3

5. Since M (x, y) = —y/x2, and N (x, y) = 1/x + 1/y,
∂M 1 ∂N
=— 2 = .
∂y x ∂x




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