Question

The transport of a substance across a capillary wall in lung physiology has been modeled by the differential equation dh/dt=(-R/v)(h/(k+h)) where is the hormone concentration in the bloodstream, is time, R is the maximum transport rate, V is the volume of the capillary, and is a positive constant that measures the affinity between the hormones and the enzymes that assist the process. Solve this differential equation to find a relationship between h and t.

Answer 1

Answer:

$k~\text{log}h + h = \frac{-R(t+c)}{v}$ is the solution to the given differential equation.  

Step-by-step explanation:

We are given the following information in the graph:

$\displaystyle\frac{dh}{dt} = \bigg(\frac{-R}{v}\bigg)\bigg(\frac{h}{k+h}\bigg)$

where h is the hormone concentration in the bloodstream, R is the maximum transport rate, t is time, v is the volume of the capillary, and k is a positive constant that measures the affinity between the hormones.

We have to solve the given differential equation:

$\displaystyle\frac{dh}{dt} = \bigg(\frac{-R}{v}\bigg)\bigg(\frac{h}{k+h}\bigg)\\\\\bigg(\frac{-v}{R}\bigg)\bigg(\frac{k+h}{h}\bigg) dh = dt\\\\\bigg(\frac{-v}{R}\bigg)\bigg(\frac{k}{h} + 1\bigg) dh = dt\\\\\text{Integrating both sides}\\\\\int \bigg(\frac{-v}{R}\bigg)\bigg(\frac{k}{h} + 1\bigg) dh = \int dt\\\\\bigg(\frac{-v}{R}\bigg)\bigg(k~\text{log}h + h\bigg) = t + c\\\\\text{where c is the constant of integration}\\\\k~\text{log}h + h = \frac{-R(t+c)}{v}$