Let us solve the given problem step-by-step:

Problem:

The rate of change in the concentration $C(t)$ of a drug with respect to time $t$ is given by:

$\frac{dC}{dt} = -kC + D(t),$

where $D(t)$ is the dosage at time $t$, and $k$ is the rate at which the drug leaves the bloodstream.

Part (a): Solve the linear equation to show that if $C(0) = 0$, then:

$C(t) = e^{-kt} \int_0^t e^{ky} D(y) \, dy.$

Solution:

This is a first-order linear differential equation. The general method for solving such equations involves finding an integrating factor.

1. Rewrite the equation: $\frac{dC}{dt} + kC = D(t).$

2. Find the integrating factor: The integrating factor is: $\mu(t) = e^{\int k \, dt} = e^{kt}.$

3. Multiply through by the integrating factor: $e^{kt} \frac{dC}{dt} + k e^{kt} C = e^{kt} D(t).$

   The left-hand side is the derivative of $e^{kt} C$: $\frac{d}{dt} \left( e^{kt} C \right) = e^{kt} D(t).$

4. Integrate both sides: $e^{kt} C = \int e^{kt} D(t) \, dt.$

5. Solve for $C(t)$: Divide through by $e^{kt}$: $C(t) = e^{-kt} \int e^{kt} D(t) \, dt.$

   If $C(0) = 0$, the integral's lower limit is $0$, so: $C(t) = e^{-kt} \int_0^t e^{ky} D(y) \, dy.$

Thus, we have proved the desired result.

Part (b): Show that if $D(t)$ is constant $D(t) = D_0$, then:

$C(t) = \frac{D_0 (1 - e^{-kt})}{k}.$

Solution:

1. Substitute $D(t) = D_0$ into the integral: Using the result from part (a): $C(t) = e^{-kt} \int_0^t e^{ky} D_0 \, dy.$

   Since $D_0$ is a constant, it can be factored out of the integral: $C(t) = e^{-kt} D_0 \int_0^t e^{ky} \, dy.$

2. Solve the integral: The integral of $e^{ky}$ is: $\int e^{ky} \, dy = \frac{1}{k} e^{ky}.$

   Evaluating from $0$ to $t$: $\int_0^t e^{ky} \, dy = \frac{1}{k} \left( e^{kt} - e^{k \cdot 0} \right) = \frac{1}{k} \left( e^{kt} - 1 \right).$

3. Substitute back: $C(t) = e^{-kt} D_0 \cdot \frac{1}{k} \left( e^{kt} - 1 \right).$

4. Simplify: Multiply through: $C(t) = \frac{D_0}{k} \left( 1 - e^{-kt} \right).$

Thus, we have proved the desired result.

Summary of Results:

1. If $C(0) = 0$, then: $C(t) = e^{-kt} \int_0^t e^{ky} D(y) \, dy.$

2. If $D(t) = D_0$, then: $C(t) = \frac{D_0}{k} \left( 1 - e^{-kt} \right).$

Would you like further clarification on any steps? Here are five questions for practice or further exploration:

1. What is the significance of the integrating factor in solving linear differential equations?
2. How would the solution change if $C(0) \neq 0$?
3. Can you solve this equation for other forms of $D(t)$, such as $D(t) = e^{-mt}$?
4. Why does $C(t)$ approach a steady-state value as $t \to \infty$ when $D(t) = D_0$?
5. How does the parameter $k$ affect the rate at which the drug leaves the bloodstream?

Tip: Always double-check the units in your differential equations, especially in biological and physical applications, to ensure consistency in your solution.