Money in a bank account grows continuously at an annual rate of 4%. Suppose $10,000 is put into an account at time t = 0. If B is the balance in the account after t years, what is the differential equation for B and its solution?
A) ;
B) ;
C) ;
D) ;

Question

Money in a bank account grows continuously at an annual rate of 4%. Suppose $10,000 is put into an account at time t = 0. If B is the balance in the account after t years, what is the differential equation for B and its solution?
A) ; 
B) ; 
C) ; 
D) ;

Quizplus · Accepted Answer

Let B(t) be the balance in the account after t years.
The rate of change of B is proportional to the current value of B with proportionality constant 0.04. This can be expressed as:

dB/dt = 0.04B

This is a separable differential equation that can be solved by separating variables and integrating:

1/B d(B) = 0.04 dt
ln(B) = 0.04t + C
B = e^(0.04t+C)

Since B(0) = 10000, we can solve for C:

10000 = e^C
C = ln(10000)

So the solution to the differential equation is:

B(t) = 10000e^(0.04t+ln(10000))
= 10000e^(ln(10000))e^(0.04t)
= 10000(1.04)^t

Money in a Bank Account Grows Continuously at an Annual