A company deposits $5,000 into an account that earns interest. The rate at which the value changes is given by dA/dt=0.0225A, where A(t) is the account value, in dollars, after t years. What will be the value of the account after 10 years if no additional deposits or withdrawals are made?

  $6261.61

Step-by-step explanation:

The solution to the differential equation is the exponential function ...

  A(t) = 5000e^(0.0225t)

We want the account value after 10 years:

  A(10) = 5000e^(0.225) = 6261.61

The value of the account after 10 years will be $6,261.61.

The rate of change equation basically tells you that interest is compounded continuously. After working interest problems for a while you know the formula for that is the exponential formula A = A0·e^(rt).

Or, you can solve the differential equation using separation of variables:

  dA/A = 0.0225dt

  ln(A) = 0.0225t +C . . . . integrate

  A(t) = A0·e^(0.0225t) = 5000·e^(0.0225t) . . . . solution for A(0) = 5000