Question

Differential equations have been used extensively in the study of drug dissolution for patients given oral medications. One such equation is the Weibull equation for the concentration c(t) of the drug: dc/dt = k/tᵇ (c_s - c) where k and c_s are positive constants and 0 < b < 1. Verify that c(t) = c_s(1 - e^-alpha t^1 - b) is a solution of the Weibull equation for t > 0, where a = k/(1 - b) is a solution of the Weibull equation for t > 0, where a = k/(1 - b). What does the differential equation say about how drug dissolution occurs?

Answer 1

To confirm that the given function is a solution to the Weibull equation, its derivative is calculated and shown to match the Weibull equation's right side. The differential equation suggests that drug dissolution occurs more rapidly at first and gradually slows down, indicating how drug concentrations change over time in the body.

Step-by-step explanation:

To verify that c(t) = c_s(1 - e-alpha t1 - b) is a solution to the Weibull equation dc/dt = k/tb (c_s - c), we need to compute the derivative of c(t) with respect to t and show that it matches the right side of the Weibull equation.

First, calculate the derivative of c(t):
dc/dt = c_s × alpha × (1-b) × t-(b) × e-alpha t1 - b

Now, substitute alpha with k/(1-b) in the derivative:
dc/dt = c_s × k / (1-b) × (1-b) × t-(b) × e-alpha t1 - b
= k/tb × c_s × (1 - e-alpha t1 - b)

The given solution matches the Weibull equation when rearranged to:
dc/dt = k/tb × (c_s - c_s(1 - e-alpha t1 - b))
= k/tb × (c_s - c)

This proves that the given function c(t) is a valid solution to the Weibull equation.

The differential equation implies how drug dissolution occurs by indicating that the rate of change of drug concentration over time depends inversely on the power of time, influenced by constants k and c_s. The dissolution occurs faster initially and slows down as time passes, reaching a saturation point where the concentration is close to c_s.