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

Answer 1

Answer:

$\bf x(t)=10(1-e^{-0.005*t})$

Step-by-step explanation:

The differential equation

$\bf \displaystyle\frac{dx}{dt}=0.005(10-x)$

can be solved by separation of variables. Write the equation as

$\bf \displaystyle\frac{dx}{10-x}=0.005dt$

Integrate on both sides

$\bf \int\displaystyle\frac{dx}{10-x}=\int0.005dt\Rightarrow -ln(10-x)=0.005t+C\Rightarrow\\\\\Rightarrow ln(10-x)^{-1}=0.005t+C\Rightarrow (10-x)^{-1}=e^{0.005t}e^C$

where C is a constant.

$\bf e^C$ is also a constant and we will keep calling it C, (there is no reason to change the letter). We have then

$\bf (10-x)^{-1}=Ce^{0.005t}\Rightarrow \displaystyle\frac{1}{10-x}=Ce^{0.005t}\Rightarrow 10-x=\displaystyle\frac{1}{Ce^{0.005t}}\Rightarrow\\\\\Rightarrow x(t)=10-(1/C)e{-0.005t}$

(1/C) is a constant, and for the same reason we will keep calling it C. So the general solution is

$\bf x(t)=10-Ce^{-0.005t}$

Now, we use the initial condition x(0)=0

$\bf x(0)=10-Ce^{-0.005*0}=0\Rightarrow C=10$

and the particular solution is

$\bf x(t)=10-10e^{-0.005*t}=10(1-e^{-0.005*t})\\\\\boxed{x(t)=10(1-e^{-0.005*t})}$