Question

8.) 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

This is a separable differential equation so can be written as

$\dfrac{dA}{A}=0.02\,dt$

Integrating, then taking antilogs, you have

$\displaystyle\int{\frac{dA}{A}}=0.02\int{dt}\\\\\ln(A)+C=0.02t\\\\A=C\cdot e^{0.02t}$

The exponential factor is 1 at t=0, so the multiplying constant is 4000.

$A=4000e^{0.02t}$