Question

8r.) Find the amount A in an account after t years given the following conditions.

Answer 1

dA/dt = 0.02A

dA/A = 0.02dt

integrating left n right

lnA = 0.02t + C where C is a constant

at t=0, A=4000

C=ln4000

lnA = 0.02t + ln4000

0.02t = lnA-ln4000

t = ln(A/4000) / 0.02

t = 40ln(A/4000)

Answer 2

You can rewrite the differential equation as

`50\cdot(dA)/(A)=dt\\\\50ln((A))+C=t\qquad\mbox{integrating}\\\\t=50\ln{\left((A)/(4000)\right)}\qquad\mbox{applying the boundary condition}`

This last expression looks like the one you describe. If you solve for A instead of t, you get

`A=4000\cdot e^(0.02t)`

This is the same as your other answer.