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MATH1013 Quizzes

Quiz 6: Systematic solution of first Order Linear Difference Equations
Question 1 Questions
Solve the first order linear difference equation
Xn Xn1 = 0.1Xn1
where X1 = 100.
a)
Xn = 100(1.1)n1.
 b)
Xn = 100(0.9)n1
c)
Xn = 100(1.1)n
 d)
Xn = 0.9(100)n

Choice (a) is correct!
Xn = 1.1Xn1 so the solution is Xn = (1.1)n1X1 = 100(1.1)n1.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Solve the first order linear difference equation
Xn = 4 3Xn1 400
where X1 = 1000.
a)
Xn = 2600 3 (4 3)n1 + 400 3
 b)
Xn = 2600 3 (4 3)n1 400 3
c)
Xn = 200(4 3)n1 1200
 d)
Xn = 1200 200(4 3)n1

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Xn = 4 3n1X 1 400 k=0n2 4 3k . Now k=0n2 4 3k = 4 3 n1 1 4 3 1 = 3 4 3n1 1, soXn = 1000 4 3n1 1200 4 3n1 1 = 1200 200 4 3n1.
A disease spreads through a population in the following way. Each year 1000 new cases of the disease occur and half of the existing cases are cured. At the beginning of the year 2000 there were 1200 cases of the disease. Find the difference equation which describes this situation and solve it.
a)
The difference equation is Xn = 1.5Xn1 + 100 and the solution is
Xn = 3200(1.5)n1 200.
 b)
The difference equation is Xn = 1.5Xn1 + 100 and the solution is
Xn = 2000 800(1.5)n1.
c)
The difference equation is Xn = 0.5Xn1 + 100 and the solution is
Xn = 3200(0.5)n1 200.
 d)
The difference equation is Xn = 0.5Xn1 + 100 and the solution is
Xn = 2000 800(1.5)n1.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Xn = 0.5Xn1 + 1000, Xn = (0.5)n1X 1 + 1000 k=0n2(0.5)k . Now k=0n2(0.5)k = (0.5)n1 1 0.5 1 = 2((0.5)n1 1), soXn = 1200(0.5)n1 2000((0.5)n1 1) = 2000 800(0.5)n1.
A population increases by 20% per annum and every year 150 are culled. If the initial population size is 100, how long does it take for the population to reach 1500 ?
a)
At the beginning of the 6th year.
 b)
At the beginning of the 7th year.
c)
At the beginning of the 9th year.
 d)
Never, the population is decreasing.

Choice (a) is incorrect
Choice (b) is correct!
Xn = 1.2Xn1 150, Xn = (1.2)n1X 1 150 k=0n2(1.2)k . Now k=0n2(1.2)k = (1.2)n1 1 1.2 1 = 5((1.2)n1 1), soXn = 1000(1.2)n1 750((1.2)n1 1) = 750 + 250(1.2)n1 Nowsolving1500 = 750 + 250(1.2)n1 250(1.2)n1 = 750, (1.2)n1 = 750 250 = 3, n 1 = ln3 ln1.2 = 6.02 n = 7.02. Therefore the population will be 1500 at the beginning of the 7th year.
Choice (c) is incorrect
Choice (d) is incorrect
Find the general solution to the difference equation
xn+1 9xn1 = 0.
a)
xn = 9n1x1
 b)
xn = A3n + An3n
c)
xn = (A + B(1)n)3n
 d)
xn = A3n

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The auxiliary equation is λ2 9 = (λ 3)(λ + 3) = 0, so the solution is
xn = A3n + B(3)n = (A + B(1)n)3n.
Choice (d) is incorrect
Find the general solution to the difference equation
xn+1 = 4xn1 + 21xn.
a)
xn = A(7n) + B(3n)
 b)
xn = A(7n) + B(3n)
c)
xn = A(7n) + B(3n)
 d)
xn = A(7n) + B(3n)

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The difference equation is xn+1 4xn1 21xn = 0. The auxiliary equation is λ2 4λ 21 = (λ 7)(λ + 3) = 0, so the solution is
xn = A(7n) + B(3)n.
Choice (d) is incorrect
Find the solution to the second order linear difference equation
xn+2 3xn+1 4xn = 0,
where x0 = 1 and x1 = 2.
a)
xn = A4n + B(1)n
 b)
xn = A(4)n + B
c)
xn = (0.6)4n + (0.4)(1)n
 d)
xn = (0.2)(4)n + 1.2

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The auxiliary equation is λ2 3λ 4 = (λ 4)(λ + 1) = 0, so the solution is
xn = A4n + B(1)n.
Using the initial conditions A + B = 1and4A B = 2 impliesA = 0.6andB = 0.4.
Choice (d) is incorrect
A population increases in such a way that its growth in any year is three times its growth in the previous year. Find the second order linear difference equation that models this population.
a)
xn+2 4xn+1 + 3xn = 0
 b)
xn+2 + 2xn+1 + 3xn = 0
c)
xn+2 3xn = 0
 d)
xn+2 2xn+1 + 3xn = 0

Choice (a) is correct!
xn+2 xn+1 is the growth in year n + 2. xn+1 xn is the growth in the previous year i.e. year n + 1. Hence xn+2 xn+1 = 3(xn+1 xn) and xn+2 4xn+1 + 3xn = 0.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
A population increases in such a way that its growth in any year is three times its growth in the previous year. Find the population in year n if the initial population was 1000 and the population the next year was 1400.
a)
xn = 600(3n) + 400(1)n
 b)
xn = 200(3n) + 800
c)
xn = 400 3 (4n) + 2600 3
 d)
xn = 80(4n) + 920(1)n

Choice (a) is incorrect
Choice (b) is correct!
From the previous question we know that the difference equation is xn+2 4xn+1 + 3xn = 0. Hence the auxiliary equation is λ2 4λ + 3 = (λ 3)(λ 1) = 0. Therefore the solution is
xn = A(3)n + B(1)n.
Using initial conditions A + B = 1000and3A + B = 1400 impliesA = 200andB = 800. Hence xn = 200(3)n + 800.
Choice (c) is incorrect
Choice (d) is incorrect
During an epidemic the number of new cases of a disease occurring during the nth week is equal to twice the number of cases that existed at the end of the n 2th week. Find the number of cases in the nth week if x0 = x1 = 1.
a)
xn = (1 2 + 2 4 )(2)n + (1 2 2 4 )(2)n
 b)
xn = (1 2 + 2 4 )(2)n + (1 2 2 4 )(2)n
c)
xn = 2 3(2n) + 1 3(1)n
 d)
xn = 1 2(2n) + 1 2

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Since xn xn1 is the number of new cases occurring in week n we have
xn xn1 = 2xn2 xn xn1 2xn2 = 0.
Hence the auxiliary equation is λ2 λ 2 = (λ 2)(λ + 1) = 0 so the solution is
xn = A(2)n + B(1)n.
Using initial conditions A + B = 1and2A B = 1 impliesA = 2 3andB = 1 3. Hence xn = 2 3(3)n + 1 3(1)n.
Choice (d) is incorrect