Differential Equations and Integrals

The video tutorial explains how to model a bank account balance using a differential equation. Starting with an initial balance of $80,000, the account earns 8% interest annually, with an additional $4,000 deposited each year. The tutorial demonstrates solving the differential equation using separation of variables, integration, and substitution. It then converts the equation to an exponential form to solve for the account balance over time. Finally, the tutorial uses the initial condition to find the particular solution, determining the constant 'A' to complete the solution.

What is the initial amount in the bank account?

• $8,000

• $4,000

• $40,000

• $80,000

What is the annual interest rate applied to the bank account?

• 12%

• 5%

• 10%

• 8%

Which method is used to solve the differential equation in this problem?

• Separation of variables

• Laplace transform

• Integration by parts

• Partial fraction decomposition

What substitution is used during the integration process?

• w = 0.08y + 4,000

• z = 0.08y + 4,000

• v = 0.08y + 4,000

• u = 0.08y + 4,000

What is the integral of 1/u with respect to u?

• ln|u|

• u^2/2

• e^u

• 1/u

How is the logarithmic equation converted to an exponential form?

• By taking the square root

• By differentiating both sides

• By integrating both sides

• By exponentiating both sides

What is the general solution form of the differential equation?

• y = A * e^(0.08t) - 4,000/0.08

• y = A * e^(0.08t) + 4,000/0.08

• y = A * e^(0.08t) - 4,000

• y = A * e^(0.08t) + 4,000

What is the value of the constant 'A' in the particular solution?

• 8,000

• 10,400

• 6,400

• 4,000

What is the particular solution for Y(t) given the initial condition?

• Y(t) = 10,400 * e^(0.08t) + 4,000/0.08

• Y(t) = 10,400 * e^(0.08t) - 4,000/0.08

• Y(t) = 10,400 * e^(0.08t) + 4,000

• Y(t) = 10,400 * e^(0.08t) - 4,000

What is the purpose of using the initial condition in solving the differential equation?

• To determine the constant 'A'

• To find the interest rate

• To calculate the time period

• To adjust the deposit amount