DSC3707 Sol Ass03 S2-2022_12c21ab2cd96bb6ee0ff55a995082dc3
pdf
keyboard_arrow_up
School
University of South Africa *
*We aren’t endorsed by this school
Course
3707
Subject
Mathematics
Date
Nov 24, 2024
Type
Pages
4
Uploaded by ChancellorFang8501
DSC3707: Solutions for Assignment 03, Semester 2, 2022
Question 1
Determine the general solution to the recurrence equation.
1.1
y
t
-
5
y
t
-
1
+ 6
y
t
-
2
= 0
1.2
y
t
+ 2
y
t
-
1
+
y
t
-
2
= 0
1.3
y
t
-
3
y
t
-
1
+ 9
y
t
-
2
= 0
Solution:
1.1
y
t
-
5
y
t
-
1
+ 6
y
t
-
2
= 0. The auxiliary equation is
z
2
-
5
z
+ 6 = (
z
-
2)(
z
-
3) = 0 with solution
z
= 2 and
z
= 3 The homogeneous equation
y
t
-
5
y
t
-
1
+ 6
y
t
-
2
= 0
therefore has the general solution
y
2
=
A
2
t
+
B
3
t
.
1.2
y
t
+ 2
y
t
-
1
+
y
t
-
2
= 0. The auxiliary equation is
z
2
+ 2
z
+ 1 = (
z
+ 1)
2
= 0 with solution
z
=
-
1
.
The
homogeneous equation
y
t
+ 2
y
t
-
1
+
y
t
-
2
= 0 therefore has the general solution
y
t
= (
Ct
+
D
)(
-
1)
t
.
1.3
y
t
-
3
y
t
-
1
+ 9
y
t
-
2
= 0 The auxiliary equation is
z
2
-
3
z
+ 9 = 0 has no real solutions since
a
1
=
-
3 and
a
2
= 9 implies that
a
2
1
-
9
a
2
= (
-
3)
2
-
9(9) =
-
72
<
0
.
We use the formulae on p 290:
p
=
-
a
1
2
= 1
, r
=
√
a
2
=
√
9 = 3
.
Then cos
θ
=
p
r
=
1
2
and hence
θ
=
π
3
= 60 degree. The homogeneous equation
y
t
-
3
y
t
-
1
+ 9
y
t
-
2
= 0
therefore has the general solution
y
t
=
A
3
t
cos
θt
+
β
3
t
sin
θt
Question 2
Income from an investment can be described by the second-order recurrence
y
t
-
3
y
t
-
1
-
4
y
t
-
2
= 6
.
2.1 Determine the general solution to the recurrence equation.
2.2 Determine the specific solution that satisfies the initial conditions
y
0
= 1
, y
1
=
-
1
.
Solution:
2.1 The auxiliary equation is
z
2
-
3
z
-
4 = (
z
+1)(
z
-
4) = 0 with solution
z
=
-
1 and
z
= 4. The homogeneous
equation
y
t
-
3
y
t
-
1
-
4
y
t
-
2
= 0
therefore has the general solution
y
t
=
A
(
-
1)
t
+
B
(4)
t
.
A particular solution of the non-homogeneous equation is the constant solution
y
*
=
-
6
1
-
3
-
4
=
-
1
Which solves
y
* -
3
y
*
= 6. Hence the general solution is
y
t
=
-
1 +
A
(
-
1)
t
+
B
(4)
t
.
2.2 Since
y
0
= 1, we have
A
+
B
-
1 = 1 while
Y
1
=
-
1 yields
A
(
-
1) =
B
(4)
-
1 =
-
1 Solving these two
equations simultaneously yields
B
=
2
5
and
A
=
8
5
.
Question 3
Consider the demand function
q
D
(
p
) = 5
-
p
ln(
p
)
,
p
≥
0
,
10
where
p
is the price per unit and
q
the number of units per hour. Determine the gradient of this demand function.
Solution:
The gradient is given by
d
dp
(5
-
p
ln
p
) = 0
-
d
dp
(
p
ln
p
)
=
-
ln
p
-
1
.
Question 4
Consider the demand function
q
D
(
p
) =
8 000
p
2
+ 1
4.1 For which values of
p
will the demand be equal to 1560?
4.2 Determine the
R
8 000
p
2
+1
dp.
Solution:
4.1 We solve the equation
1 560 =
8 000
p
2
+ 1
,
or
1 560
p
2
+ 1 560 = 8 000
Dividing through by 40 we get
39
p
2
= 161
p
=
q
161
39
≈
2
,
032
,
ignoring then negative value.
4.2
R
8 000
p
2
+1
dp
= 8 000
R
1
p
2
+1
dp.
Use maxima or note that the
R
1
p
2
+1
pd
is a standard integra l and
R
1
p
2
+1
pd
= arctan
P
, where arctan
p
is
the inverse of tan
p
, also denoted by tan
-
1
p
.
Question 5
The inverse demand function for a good is
p
D
(
q
) =
192
q
2
+ 4
q
+ 3
,
and the equilibrium price is
p
*
= 4. Determine the equilibrium quantity and the consumer surplus.
Solution:
p
D
(
q
) =
192
q
2
+ 4
q
+ 3
,
Equilibrium quantity
192
q
2
+ 4
q
+ 3
= 4
→
192 = 4
(
q
2
+ 4
q
+ 3
)
with roots
q
= 5 and
q
=
-
9.
CS
= 192
R
5
0
1
q
2
+4
q
+3
dq
-
(4
×
5)
= 192
R
5
0
1
(
q
+1)(
q
+3)
dq
-
20
.............
1
(
q
+3)(
q
+1)=
A
q
+3
+
B
q
+1
= 192
R
5
0
1
2
(
q
+1)
-
1
2
(
q
+3)
dq
-
20
= 96
R
5
0
1
(
q
+1)
-
1
(
q
+3)
dq
-
20
= 96 [ln(
q
+ 1)
-
ln(
q
+ 3)]
5
0
-
20 = 57
,
8493
.
Page 2
Question 6
The elasticity of demand for a good is
ε
(
p
) =
2
p
2
p
2
+ 1
.
Given that
q
= 4 when
p
= 1, determine the demand function
q
D
(
p
)
.
Solution:
ε
(
p
) =
2
p
2
p
2
+ 1
.
We have
-
p
q
dq
dp
=
2
p
2
p
2
+ 1
,
where
q
=
q
D
(
p
) is the demand function. This may be written as a separable differential equation
dq
dp
=
-
2
p
p
2
+ 1
q.
Separating and integrating
Z
1
q
dq
=
Z
-
2
p
p
2
+ 1
dp.
Using substitution
r
=
p
2
+ 1
ln
q
=
-
ln(
p
2
+ 1) +
c,
which is valid since
q
and
P
2
+ 1 are positive. Taking the exponential of both sides
q
=
e
c
e
-
ln(
p
2
+1)
=
e
c
p
2
+ 1
=
K
p
2
+ 1
,
where
K
=
e
c
is some positive constant. Given that
q
(1) = 4
,
we have 4 =
K
2
, so
K
= 8 Therefore the demand
function is
q
D
(
p
) =
8
p
2
+ 1
.
Question 7
Suppose that consumer demand depends upon the price-trend according to the formula
q
= 12
-
p
-
3
dp
dt
+
d
2
p
dt
2
.
If the supply function
q
s
(
p
) =
-
8 + 3
p
, write down the condition for equilibrium and determine the equilibrium price
p
(
t
) when
p
(0) = 7 and
p
0
(0) = 9.
Solution:
The condition for equilibrium is that supply matches demand. Hence
-
8 + 3
p
= 12
-
p
-
3
dp
dt
+
d
2
p
dt
2
.
So we get
-
20 =
-
4
p
-
3
dp
dt
+
d
2
p
dp
2
.
A particular solution is
p
= 5. The auxiliary equation is
z
2
-
3
z
-
4 = 0
or
(
z
-
4)(
z
+ 1) = 0
,
so
z
= 4 or
z
=
-
1. Hence the general solution is
p
(
t
) = 5 +
Ae
4
t
+
Be
-
t
.
Page 3
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Using
p
0
(0) = 9 and
p
(0) = 7 gives the equations
7 = 5 +
A
+
B
9 = 4
A
-
B.
Then
A
=
11
5
;
B
=
-
1
5
and the equilibrium price is
p
(
t
) = 5 +
11
5
e
4
t
-
1
5
e
-
t
.
Page 4