A drug is administered intravenously at a constant rate of 2

A drug is administered intravenously at a constant rate of 2 mg/hour and excreted at a rate proportional to the present quantity, Q(t), with constant of proportionality k = 0.3. Find the differential equation for the quantity, Q(t), in milligrams, of the drug in the body at time t hours.

Solution

so according to the problem the equation we get is dQ/dt = r - k*Q where r = 2 mg/hr and k=0.3

Rewrite this as:

dQ/dt = -k*(Q - r/k)

dQ/(Q - r/k) = -k dt

ln[(Q - r/k)] + C = -k*t

ln[(Q - r/k)] = -(C+k*t)

[(Q - r/k)] = e^-(C+k*t)

Q(t) = r/k + C*e^-k*t here taking e^-(C+k*t) = (e^-C * e^{-kt )} and e^-C as also a constant C

If Q(0) = 0, then:

0 = r/k + C*e⁽⁰⁾ = r/k + C

C = -r/k

so

Q(t) = (r/k)*(1 - e^(-k*t))

substituting values of r and k we get,

Q(t) = (2/0.3)*(1- e^-0.3t) = 6.67*(1- e^-0.3t) in milligramrs at time t hrs

The steady-state concentration is given by Q(infinity) = r/k= 2/0.3=6.67 mg/hr