# differential equation

If P(t) is the amount of dollars in a savings bank account that pays a yearly interest rate of r% compounded continuously ,then dP/dt=(r/100)(P) , t in years . Assume the interest is 5% annually ,P(0)=\$1000 ,and no monies are withdrawn a)how much will be in the account after 2 years ? b)when will the account reach \$4000? c)if \$1000 is added to the account every 12 months, how much will be in the account after 3.1/2 year ?

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1. dP/dt = .05 P
dP/P = .05 dt
ln P = .05 t

P = Pi e^.05t
if Pi = 1000 and t = 2
P = 1000 e^(.1) = 1105.17

4000 = 1000 e^.05t
e^.05 t = 4
.05 t = ln 4
t = 27.72 years

1000 for 1 year = 1051
2051 for year 2 = 2156
3156 for year 3 = 3318
4318 * e^.025 =4427

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2. Last line in answer has an error, it should say:
3318*e^.025=3402.27

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