a bank account earns 7% annual interest compounded continuously. you deposit $10,000 in the account, and withdraw money continuously from the account at a rate of $1000 per year.

a. write the differential equation for the balance, B, in the account after t years
b. what is the equilibrium solution to the differential equation?
c. find the solution to the differential equation.
d. how much is in the account after 5 years?
e. graph the solution. what happens to the balance in the long run?

How long

a. To write the differential equation for the balance, B, in the account after t years, we need to consider the factors that affect the balance over time. In this case, there are two factors: the continuous interest and the continuous withdrawals.

Let's break it down:

1. Continuous interest: The bank account earns 7% annual interest compounded continuously. This means that the balance increases continuously at a rate proportional to the balance at any given time.

2. Continuous withdrawals: Money is being continuously withdrawn from the account at a rate of $1000 per year. This means that the balance decreases continuously at a constant rate.

Combining these two factors, the differential equation for the balance, B, can be written as:

dB/dt = rB - w

Where dB/dt is the rate of change of the balance with respect to time (t), r is the annual interest rate (0.07), B is the balance, and w is the constant withdrawal rate ($1000 per year).

b. To find the equilibrium solution to the differential equation, we need to set the rate of change of the balance equal to zero. This means that the balance is not changing over time. In other words, the equilibrium solution occurs when the interest earned is equal to the amount withdrawn.

Setting the differential equation equal to zero:

0 = rB - w

Solving for the balance, B:

B = w/r

Therefore, the equilibrium solution to the differential equation is B = $1000/0.07 = $14,285.71.

c. To find the solution to the differential equation, we need to solve the differential equation using separation of variables.

Separate the variables and integrate:

1/(B - w) dB = r dt

Integrating both sides:

ln|B - w| = rt + C

Where C is the constant of integration.

Using the initial condition that B = $10,000 when t = 0, we can solve for C:

ln|10000 - 1000| = r(0) + C
ln|9000| = C

Therefore, the solution to the differential equation is:

ln|B - w| = rt + ln|9000|

d. To find the balance in the account after 5 years, we substitute t = 5 into the solution and solve for B:

ln|B - 1000| = 0.07(5) + ln|9000|
ln|B - 1000| = 0.35 + ln|9000|

Now, we can exponentiate both sides:

|B - 1000| = e^(0.35 + ln|9000|)
|B - 1000| = e^0.35 * 9000

Taking the positive root:

B - 1000 = e^0.35 * 9000
B = e^0.35 * 9000 + 1000

Using a calculator, we find that B ≈ $14,631.54.

e. To graph the solution, we can plot the balance, B, on the y-axis and the time, t, on the x-axis. The balance initially starts at $10,000 and gradually increases due to the continuous interest, while also decreasing due to the continuous withdrawals.

Over time, as t approaches infinity, the balance eventually approaches the equilibrium solution of $14,285.71. This means that in the long run, the balance will stabilize at around $14,285.71, as the interest earned offsets the continuous withdrawals.