Adeposit of $5000 is made to a bank account paying 1.5% annual interest, compounded continuously. (a) write the differential equation for the balance in the account, b, as a function of time, t, in years. (b) solve the differential equation. (c) how much money is in the account in 10 years?

Please find the detailed answer as follows:

Explanation:

A. Step 1. The differential equation is

dB / dt = 0.015B - 5000

Step 2.   now we must integrate it to find it ias a function of time

dB /0.015B - 500 = dt

Step 3. integrate

=> ln ( 0.015 B - 5000) = 0.015t + 0.015c

=> 0.015B - 5000 = e^0.015t .e^0.015c

here .e^0.015c is constant let it be Bo

b. =>B = 1/0.015 [ Bo e^0.015t + 5000 ]

at t = 0 , B = 5000  

=> 5000 x 0.015 = Bo => B o = 75

c . B = 1/0.015 [ 75 e^0.015t + 5000 ]

put t = 10  

=> B = $ 339142.50

answer; swot analysis;