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.