Answer
a) $dx/dt=500-0.4x$
b) $1250 mg$ is present after a long time
Work Step by Step
a)A fluid of concentration $5mg/(cm^3)$ isn going into the patient. It is going into them at a rate of $100(cm^3)/hour$. Therefore, the delivery rate of the drug is $(\frac{5mg}{cm^3})*\frac{100cm^3}{1 hour}=\frac{500 mg}{1 hour}$. The drug is leaving the patient at a rate "proportional to the amount present" by 0.4; this means that the drug is exiting the body at a rate of $0.4x$ where $x$ is the concentration of the drug at a particular time. A differential equation may represent this scenario as $\frac{∆concentration}{∆time}$ or $\frac{dx}{dt}$ where $t$ is time. The change in concentration of the drug with respect to time is the sum of all inputs and outputs contributing to the concentration of the drug. In this case, there are two: $\frac{500 mg}{1 hour}$ going in and $0.4x$ $mg/hour$ exiting. So the differential equation is $dx/dt=(500-0.4x)mg/hour$.
b) after a long time, the concentration will no longer be changing because it will have reached a point where the amount going in is equal to the amount going out, i.e. $\frac{500 mg}{1 hour}$ going in and $\frac{500 mg}{1 hour}$ going out. If this is happening then $\frac{dx}{dt}=0$; plug this into the equation from a) to get $dx/dt=0=(500-0.4x)mg/hour$, therefore $500(mg/hour)=0.4(xmg)/hour$, so, $1250 mg=x$.
