MA 131 Exam 1 Prep Questions with Correct Answers Graded A+
What is the difference equation? Define each variable. - Answers Yn+1 = aYn +b , Y_0 = c
- a, b are given constants
- Y_0 is the initial condition
What is the difference equation in words? - Answers [Balance] = [Previous balance] + [Interest for
period] + [Payment/Deposit]
What is the solution to the difference equation when a = 1? - Answers Yn = Y_0 + bn
What is the solution to the difference equation when a does not = 1? - Answers Yn = [b/(1 - a) + (Y_0 -
(b/(1 - a)]a^n
Suppose that a savings account contains $50 and earns 4% interest, compounded annually. At the end of
each year, a $3 withdrawal is made. Determine a formula (difference equation) which describes how to
compute each year's balance from the previous year's balance. - Answers 1) Write a word equation to
identify your variables.
[Balance] = [Previous balance] + [Interest] - [Withdrawal]
[Previous balance] = Y_0 = 50
[Interest] = 0.04
[Withdrawal] = -3
2) Rewrite it as:
Yn+1 = Yn + aYn - W
Yn+1 = Yn + (0.4)Yn - 3
Yn+1 = (1.04)Yn - 3 , Y_0 = 50
Suppose that a savings account contains $500 and earns 6% interest compounded annually. At the end
of the year a $25 withdrawal is made.
a) Determine a difference equation that describes each year's balance.
b) Solve the difference equation.
c) Find the balance after 100 years. - Answers a)
,1) Write a word equation to identify your variables.
[Balance] = [Previous balance] + [Interest] - [Withdrawal]
[Previous balance] = Y_0 = 500
[Interest] = 0.06
[Withdrawal] = -25
2) Rewrite as:
Yn+1 = Yn + aYn - W
Yn+1 = Yn + (0.06)Yn - W
Yn+1 = (1.06)Yn - 25 , Y_0 = 500
b)
1) Identify a, b and Y_0 and whether or not a = 1.
a = 1.06
b = -25
Y_0 = 500
2) Since a does not equal 1, use the solution:
Yn = [b/(1 - a) + (Y_0 - (b/(1 - a)]a^n
Plug in a, b and Y_0.
Yn = [-25/(1 - 1.06) + (500 - (-25/(1 - 1.06)]1.06^n
= [-25/(-.06) + (500 - (-25/-.06)]1.06^n
= [416.66 + (500 - 416.66)]1.06^n
= 416.67 + 83.33(1.06)^n
, c)
1) Plug in 100 for n.
Yn = 416.67 + 83.33(1.06)^n
= 416.67 + 83.33(1.06)^100
= 416.67 + 83.33(339.3020835)
= 416.67 + 28274.04262
= $28,690.71
Yn+1 = 1.04Yn - 3 , Y_0 = 50
Find Y_1 and Y_2. - Answers Y_1:
Y_1 = (1.04)Y_0 - 3
= (1.04)(50) - 3
= 49
Y_2:
Y_2 = (1.04)Y_1 - 3
= (1.04)(49) - 3
= 47.96
Suppose that the interest rate is 5% compounded monthly. Find a formula for the amount after n
months.
a) Find the solution to the difference equation where Y_0 is unknown, - Answers 1) To determine the
monthly rate, let:
r = yearly rate (in decimal form!)
m = number of times interest is given per year
i = r/m
What is the difference equation? Define each variable. - Answers Yn+1 = aYn +b , Y_0 = c
- a, b are given constants
- Y_0 is the initial condition
What is the difference equation in words? - Answers [Balance] = [Previous balance] + [Interest for
period] + [Payment/Deposit]
What is the solution to the difference equation when a = 1? - Answers Yn = Y_0 + bn
What is the solution to the difference equation when a does not = 1? - Answers Yn = [b/(1 - a) + (Y_0 -
(b/(1 - a)]a^n
Suppose that a savings account contains $50 and earns 4% interest, compounded annually. At the end of
each year, a $3 withdrawal is made. Determine a formula (difference equation) which describes how to
compute each year's balance from the previous year's balance. - Answers 1) Write a word equation to
identify your variables.
[Balance] = [Previous balance] + [Interest] - [Withdrawal]
[Previous balance] = Y_0 = 50
[Interest] = 0.04
[Withdrawal] = -3
2) Rewrite it as:
Yn+1 = Yn + aYn - W
Yn+1 = Yn + (0.4)Yn - 3
Yn+1 = (1.04)Yn - 3 , Y_0 = 50
Suppose that a savings account contains $500 and earns 6% interest compounded annually. At the end
of the year a $25 withdrawal is made.
a) Determine a difference equation that describes each year's balance.
b) Solve the difference equation.
c) Find the balance after 100 years. - Answers a)
,1) Write a word equation to identify your variables.
[Balance] = [Previous balance] + [Interest] - [Withdrawal]
[Previous balance] = Y_0 = 500
[Interest] = 0.06
[Withdrawal] = -25
2) Rewrite as:
Yn+1 = Yn + aYn - W
Yn+1 = Yn + (0.06)Yn - W
Yn+1 = (1.06)Yn - 25 , Y_0 = 500
b)
1) Identify a, b and Y_0 and whether or not a = 1.
a = 1.06
b = -25
Y_0 = 500
2) Since a does not equal 1, use the solution:
Yn = [b/(1 - a) + (Y_0 - (b/(1 - a)]a^n
Plug in a, b and Y_0.
Yn = [-25/(1 - 1.06) + (500 - (-25/(1 - 1.06)]1.06^n
= [-25/(-.06) + (500 - (-25/-.06)]1.06^n
= [416.66 + (500 - 416.66)]1.06^n
= 416.67 + 83.33(1.06)^n
, c)
1) Plug in 100 for n.
Yn = 416.67 + 83.33(1.06)^n
= 416.67 + 83.33(1.06)^100
= 416.67 + 83.33(339.3020835)
= 416.67 + 28274.04262
= $28,690.71
Yn+1 = 1.04Yn - 3 , Y_0 = 50
Find Y_1 and Y_2. - Answers Y_1:
Y_1 = (1.04)Y_0 - 3
= (1.04)(50) - 3
= 49
Y_2:
Y_2 = (1.04)Y_1 - 3
= (1.04)(49) - 3
= 47.96
Suppose that the interest rate is 5% compounded monthly. Find a formula for the amount after n
months.
a) Find the solution to the difference equation where Y_0 is unknown, - Answers 1) To determine the
monthly rate, let:
r = yearly rate (in decimal form!)
m = number of times interest is given per year
i = r/m
