Money differential equation problem?
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A country has 10 billion dollars in paper currency in circulation, and each day 33 million dollars 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 bank. Let x=x(t) denote the number of new dollars in circulation after t days with units in billions and x(0)=0. A. Determine a differential equation which describes the rate at which x is growing: B. Solve the differential equation subject to the initial conditions given above. x(t)= C. How many days will it take for the new bills to account for 90 percent of the currency in circulation?
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Answer:
Every day, 33 million dollars come into the banks. If x(t) is the number of new dollars (in billions) in circulation at time t, and there is a total of 10 billion dollars in circulation, then the fraction of that 33 M$ coming into banks that is old currency is equal to 1 - x(t)/10. The rate at which the old currency gets converted is then given by: dx/dt = (0.033B$/day)*(1 - x/10B$) dx/dt = -(0.0033B$/day)*(x - 10B$) Either of the above equations is an answer to part (A). This is a separable differential equation: dx/(x -10B$) = -0.0033B$/day dt Integrate both sides: ln(x - 10B$) - ln(Xo-10B$) = -(0.0033B$/day)* t where Xo is the amount of new currency in circulation at t = 0. ln((x-10B$)/(Xo-10B$)) = -(0.0033B$/day)* t (x-10B$)/(Xo-10B$) = exp( -(0.0033B$/day)* t) x(t) = 10B$ + (Xo - 10B$)*exp( -(0.0033B$/day)* t) We are told that x(0) = Xo = 0, so: x(t) = 10B$ - 10B$*exp( -(0.0033B$/day)* t) x(t) = 10B$ * [1 - exp( -(0.0033B$/day)* t)] This is the answer to part (B) of the question. We now want to know when x(t) = 9B$ (i.e., when 90% of the currency in circulation is new. Set x(t) equal to 9B$ and solve for t: 9B$ = 10B$ * [1 - exp( -(0.0033B$/day)* t)] 1 - 0.9 = exp( -(0.0033B$/day)* t) ln(0.1) = -(0.0033B$/day)* t ln(10) = (0.0033B$/day)* t t = ln(10)/(0.0033B$/day) t = 697.8 days = 1.91 years
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