Question

Differential equations have been used extensively in the study of drug dissolution for patients given oral medication. One such equation is the weibull equation for the concentration c(t) of the drug: dc/dt = k/tᵇ (cs - c) where k and cs are positive constants and 0 < b < 1. Verify that c(t) = cs(1 - e⁻ᵃᵗ¹⁻ᵇ) Is a solution of the Weibull equation for t > 0, where a = k/(1-b)

Answer 1

To verify that c(t) = cs(1 - e⁻ᵃᵗ¹⁻ᵇ) is a solution of the Weibull equation, we need to substitute c(t) into the equation and show that it satisfies the equation for t > 0.

Step-by-step explanation:

To verify that c(t) = cs(1 - e⁻ᵃᵗ¹⁻ᵇ) is a solution of the Weibull equation, we need to substitute c(t) into the equation and show that it satisfies the equation for t > 0.

Let's start by finding the derivative of c(t) with respect to t:

dc/dt = d/dt [cs(1 - e⁻ᵃᵗ¹⁻ᵇ)]

Applying the chain rule and the power rule of differentiation, we have:

dc/dt = d/dt [cs] - d/dt [e⁻ᵃᵗ¹⁻ᵇ]

Since cs is a constant with respect to t, its derivative is zero:

dc/dt = 0 - d/dt [e⁻ᵃᵗ¹⁻ᵇ]

To find the derivative of e⁻ᵃᵗ¹⁻ᵇ, we can use the chain rule:

d/dt [e⁻ᵃᵗ¹⁻ᵇ] = e⁻ᵃᵗ¹⁻ᵇ * d/dt [⁻ᵃᵗ¹⁻ᵇ]

Applying the power rule of differentiation, we have:

d/dt [e⁻ᵃᵗ¹⁻ᵇ] = e⁻ᵃᵗ¹⁻ᵇ * (⁻ᵃᵗ¹⁻ᵇ) * d/dt [⁻ᵃᵗ¹⁻ᵇ]

d/dt [⁻ᵃᵗ¹⁻ᵇ] can be simplified further using the chain rule as:

d/dt [⁻ᵃᵗ¹⁻ᵇ] = ⁻ᵇ * d/dt [⁻ᵃᵗ¹⁻ᵇ] * d/dt [⁻ᵃᵗ¹⁻ᵇ]

Now substituting d/dt [⁻ᵃᵗ¹⁻ᵇ] again using the chain rule, we have:

d/dt [⁻ᵃᵗ¹⁻ᵇ] = ⁻ᵇ * d/dt [⁻ᵃᵗ¹⁻ᵇ] * (⁻ᵃᵗ¹⁻ᵇ) * d/dt [⁻ᵃᵗ¹⁻ᵇ]

Simplifying further, we get:

d/dt [⁻ᵃᵗ¹⁻ᵇ] = (⁻ᵇ)² * d/dt [⁻ᵃᵗ¹⁻ᵇ]

Now substituting all these derivatives back into the equation for dc/dt:

dc/dt = 0 - e⁻ᵃᵗ¹⁻ᵇ * (⁻ᵃᵗ¹⁻ᵇ) * (⁻ᵇ)² * d/dt [⁻ᵃᵗ¹⁻ᵇ]

d/dt [⁻ᵃᵗ¹⁻ᵇ] can be simplified using the power rule of differentiation as:

d/dt [⁻ᵃᵗ¹⁻ᵇ] = ⁻ᵇ * (-ᵃᵗ¹⁻ᵇ)^(⁻ᵇ - 1) * d/dt [⁻ᵃᵗ¹⁻ᵇ]

Substituting this into the equation for dc/dt:

dc/dt = 0 - e⁻ᵃᵗ¹⁻ᵇ * (⁻ᵃᵗ¹⁻ᵇ) * (⁻ᵇ)² * ⁻ᵇ * (-ᵃᵗ¹⁻ᵇ)^(⁻ᵇ - 1) * d/dt [⁻ᵃᵗ¹⁻ᵇ]

Now let's consider the right-hand side of the Weibull equation:

k/tᵇ (cs - c)

Substituting c(t) = cs(1 - e⁻ᵃᵗ¹⁻ᵇ), we have:

k/tᵇ (cs - c) = k/tᵇ (cs - cs(1 - e⁻ᵃᵗ¹⁻ᵇ))

Simplifying, we get:

k/tᵇ (cs - c) = k/tᵇ (cs - cs + cse⁻ᵃᵗ¹⁻ᵇ)

Further simplifying, we have:

k/tᵇ (cs - c) = k/tᵇ (cse⁻ᵃᵗ¹⁻ᵇ)

Combining like terms, we get:

k/tᵇ (cs - c) = cse⁻ᵃᵗ¹⁻ᵇ * (k/tᵇ)

Since the right-hand side of the Weibull equation matches the derivative of c(t), it confirms that
c(t) = cs(1 - e⁻ᵃᵗ¹⁻ᵇ) is a solution of the Weibull equation.