Solutions Manual An Introduction to Difference Equations By Saber Elaydi

Solutions Manual
An Introduction to Difference Equations
By
Saber Elaydi



( All Chapters Included - 100% Verified Solutions )




1

,Chapter 1

Dynamics of First Order
Difference Equations

1.1 1.2 Exercises
1. (a) x(n + 1)= (n + 1)x(n),
 x(0) = c
n−1
Q
x(n) = (i + 1)
i=0
x(0) = cn!

(b) x(n + 1) = 3n x(n), x(0) = c

n−1
!
Y
x(n) = 3i x(0)
i=0
1+2+···+(n−1)
=3 x(0)
n(n−1)
= c3 2




(c) x(n + 1) = e2n x(n), x(0) = c

n−1
!
Y
2i
x(n) = e x(0)
i=0
2
= e0 · e2 · e2 · · · e2(n−1) c
= ce2(1+2+3+···+n−1)
2(n−1)n
= ce 2


= cen(n−1)




2

,2 Dynamics of First Order Difference Equations

n
(d) x(n + 1) = n+1 x(n), x(1) = c

n−1
!
Y i
x(n) = c
i=1
i+1
c
=
n


2. Using formula (1.2.4) we obtain the general solution.


(a)

 n n−1
X  1 n−r−1
1
y(n) = + c+ 2 +
2 r=0
2
 n  n−2 n−1
1 1 X
= c+ 2r
2 2 r=0
 n  n−2 n
1 1 2 −1
= c+
2 2 2−1
 n
1
= [c + 4(2n − 1)]
2

n
(b) y(n + 1) − y(n) = 4, y(0) = c, y(1) = c
n+1
n
Write y(n + 1) = y(n) + 4, then use (1.2.4).
n+1

n−1 n−1
" #
Y i
 X n−1 Y  i 
y(n) = c+ g(r)
i=1
i+1 r=1 i=r+1
i+1
n−1
X r + 1
c
= +4
n r=1
n
n−1
c 4X
= + (r + 1)
n n r=1
 
c 4 (n − 1)n
= + +n−1
n n 2
 
c 2
= +2 n+1−
n n


3. (a) y(n + 1) = (n + 1)y(n) + 2n (n + 1)!




3

, 1.1 1.2 Exercises 3



"n−1 # n−1
Y X n−1
Y
y(n) = (i + 1) y(0) + (i + 1)2r (r + 1)!
i=0 r=0 i=r+1
n−1
X
= n! y(0) + n(n − 1) . . . (r + 2)(r + 1)! 2r
r=0
n−1
X
= n! y(0) + n! 2r
r=0
= n! [c + 2n − 1]

(b) y(n + 1) = y(n) + en
n−1
P r en −1
y(n) = y(0) + e =c+ e−1
r=0


4. (a) Let R(n) be the number of regions created by n lines. If we add
one more line, it will meet n lines and create (n + 1) new regions.
Therefore
R(n + 1) = R(n) + (n + 1), R(1) = 2.

(b) Using formula (1.2.4), the general solution can be written as follows:

n−1 n−1 n−1
!
Y X Y
R(n) = 2 1+ 1 (r + 1)
i=1 r=1 i=r+1
n−1
X
=2+ (r + 1)
r=1
n(n + 1)
=1+
2

5. (a)
Z ∞
Γ(x) = tx−1 e−t dt, x > 0
Z0 ∞
Γ(x + 1) = tx e−t dt, u = tx , dv = e−t dt, v = −e−t
0
Z ∞
∞
= −e−t tx 0 + x tx−1 e−t dt
0
= xΓ(x)
Z ∞
∞
Γ(1) = e−t dt = −e−t 0 = 1
0




4