# Suppose you deposit 8500 dollars into a savings account earning 5 percent annual interest compounded continuously

Suppose you deposit 8500 dollars into a savings account earning 5 percent annual interest compounded continuously. To pay for all your music downloads, each year you withdraw $900 in a continuous way. Let A(t) represent the amount of money in your savings account t years after your initial deposit. (A) Write the DE model for the time rate of change of money in the account. Also state the initial condition. dA/dt=? A(0)=? (B) Solve the IVP to find the amount of money in the account as a function of time. A(t)=? (C) When will your money run out? t=? • You need to put in a small deposit of your work shown here before our interest starts.Continuous compounding involves exponential rise of amount? Oct 12, 2015 at 6:05 • i have been working through it for the past hour and cannot get it right. I believe that A(0)=8500 because that is original amount of money. However I am not sure if i should be using that in dA/dt... – Pete Oct 12, 2015 at 6:09 • maybe something like 8500e^0.05y-900t – Pete Oct 12, 2015 at 6:10 ## 1 Answer As per the notions of differential coefficient of continuous derivatives ( compounding takes place each millisecond !),with annual music download payment$M$deducted. $$\frac{dA}{dt} = r\cdot A- M$$ Integrating $$\log( r A - M) = r \, t + C$$ At start using boundary condition $$t= 0 , A =P$$ $$log \,( r p - M) = C$$ $$log\, \frac{r \, A - M }{r \,P - M} =r\, t$$ $$A = ( P- M/r) e ^{r\, t} + M/r$$$ P = 8500 ; r = 0.05 ; M=900; t = \$ no of years.

Bank balance using calculus and influence of each constant can be graphed.

• I used this method but answers are still not coming out correct.
– Pete
Oct 12, 2015 at 19:48
• For dA/dt I am getting (0.05)*(8500e^(0.05t)) according to the above and then for A(t) I get 8500e^(0.05t)-900t.
– Pete
Oct 12, 2015 at 20:01
• Also the money never ran out using this so I could not find answer for t in years
– Pete
Oct 12, 2015 at 20:02
• Not sure what that has to do with the question
– Pete
Oct 12, 2015 at 20:43
• My error, please see correction made for annual payment. Oct 12, 2015 at 21:26