 [ \frac{dC}{dt} = \text{(rate in)} - \text{(rate out)} = 4 - kC ] where ( k ) is a constant of proportionality.

The given equation is a first-order linear differential equation of the form:

\[
\frac{dC}{dt} = 4 - kC
\]

where:
- \( \frac{dC}{dt} \) represents the rate of change of \( C \) with respect to time \( t \).
- \( 4 \) is the rate of substance flowing in.
- \( -kC \) represents the rate of substance flowing out, where \( k \) is a constant of proportionality.

### Solving the differential equation:

To solve \( \frac{dC}{dt} = 4 - kC \), let's follow these steps:

#### Step 1: Rearrange the equation
\[
\frac{dC}{dt} + kC = 4
\]

This is a linear differential equation, and we can solve it using an integrating factor.

#### Step 2: Find the integrating factor
The general form of a first-order linear equation is:
\[
\frac{dC}{dt} + P(t)C = Q(t)
\]
where \( P(t) = k \) and \( Q(t) = 4 \).

The integrating factor \( \mu(t) \) is given by:
\[
\mu(t) = e^{\int P(t)\, dt} = e^{\int k \, dt} = e^{kt}
\]

#### Step 3: Multiply through by the integrating factor
Multiply both sides of the equation \( \frac{dC}{dt} + kC = 4 \) by \( e^{kt} \):
\[
e^{kt} \frac{dC}{dt} + ke^{kt}C = 4e^{kt}
\]

The left-hand side simplifies to the derivative of the product \( C e^{kt} \):
\[
\frac{d}{dt} \left( C e^{kt} \right) = 4e^{kt}
\]

#### Step 4: Integrate both sides
Integrating both sides with respect to \( t \):
\[
C e^{kt} = \int 4e^{kt} \, dt
\]
The integral of \( 4e^{kt} \) is:
\[
C e^{kt} = \frac{4}{k} e^{kt} + A
\]
where \( A \) is the constant of integration.

#### Step 5: Solve for \( C(t) \)
Now, solve for \( C(t) \) by dividing both sides by \( e^{kt} \):
\[
C(t) = \frac{4}{k} + A e^{-kt}
\]

### General solution:
The general solution to the differential equation is:
\[
C(t) = \frac{4}{k} + A e^{-kt}
\]
where \( A \) is determined by the initial condition.

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Would you like to provide initial conditions to find the particular solution, or do you have any specific questions about the solution process?

### Related questions:
1. How do we apply an initial condition to determine \( A \)?
2. What does the constant \( k \) represent physically in this equation?
3. Can this differential equation model physical phenomena like drug concentration or population growth?
4. How would the solution change if \( k \) were negative?
5. What happens to \( C(t) \) as \( t \to \infty \)?

### Tip:
When solving linear differential equations, the integrating factor method is very useful and can be applied to many types of first-order equations.

[ \frac{dC}{dt} = \text{(rate in)} - \text{(rate out)} = 4 - kC ] where ( k ) is a constant of proportionality.

Learn how to solve the first-order linear differential equation dC/dt = 4 - kC using the integrating factor method. This solution explains the step-by-step process, including key concepts like the integrating factor and the general solution. Suitable for undergraduate calculus students.

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