A certain country has $10 billion in paper currency in circulation, and each day

Question

A certain country has $10 billion in paper currency in circulation, and each day $50 million comes into the country's banks. The government decides to introduce new currency by having the banks replace old bills with new ones whenever old currency comes into the banks. Let x(t) denote the amount of new currency (in billions of dollars) in circulation at time t (in days), with x(0) = 0. Then dx/dt=(fraction of currency that is old)(0.05billion$/day)
so
dx/dt=((10-x)/10)*(0.05)=0.005(10-x)

With this info, determine x(t) by solving the differential equation and using the initial condition x(0)=0

my initial answer was 1000(1-e^-t/200) but that was incorrect

Explanation / Answer

dx/dt=((10-x)/10)*(0.05)=0.005(10-x)

=> dx/dt = 0.005*(10-x)

=> dx/(10-x) = 0.005*dt

On integrating both sides, we get

=> -log(10-x) = 0.005t + c

Given x(0) = 0

=> -log(10-0) = 0.005*0 + c

=> c = -log(10)

Therefore, c = -log(10)

-log(10-x) = 0.005t - log(10)

=> log(10) - log(10-x) = 0.005t

=> log(10/(10-x)) = 0.005t

=> 10/(10-x) = e^(0.005t)

=> (10-x)/10 = e^(-0.005t)

=> 10-x = 10e^(-0.005t)

=> x = 10 - 10e^(0.005t)

Therefore x(t) = 10*{1-e^(-0.005t)}

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