Problem 5 - Applied Calculus

Question

Find particular solutions. $$\frac{d B}{d t}=4 B-100, \quad B=20\( when \)t=0$$

Step-by-Step Solution

Answer

The particular solution is \( B(t) = 25 - 5e^{4t} \).

Step 1: Identify the type of differential equation

Observe that the given differential equation \( \frac{dB}{dt} = 4B - 100 \) is a first-order linear differential equation. It can be written in the form \( \frac{dB}{dt} - 4B = -100 \).

Step 2: Find the integrating factor

For a first-order linear differential equation given by \( \frac{dy}{dt} + P(t)y = Q(t) \), the integrating factor \( µ(t) \) is \( e^{\int P(t) \, dt} \). In this case, \( P(t) = -4 \), so integrate to get \( \int -4 \, dt = -4t \). The integrating factor is \( µ(t) = e^{-4t} \).

Step 3: Multiply through by the integrating factor

Multiply each term in the equation \( \frac{dB}{dt} - 4B = -100 \) by \( e^{-4t} \) to obtain \( e^{-4t} \frac{dB}{dt} - 4Be^{-4t} = -100e^{-4t} \).

Step 4: Simplify the left side using integrating factor property

The left side simplifies to the derivative with respect to \( t \) of \( e^{-4t}B \). So the equation becomes \( \frac{d}{dt}(e^{-4t}B) = -100e^{-4t} \).

Step 5: Integrate both sides

Integrate both sides with respect to \( t \):\[ \int \frac{d}{dt}(e^{-4t}B) \, dt = \int -100e^{-4t} \, dt \]The left side simplifies to \( e^{-4t}B \), and the right side integrates to \( 25e^{-4t} + C \). Thus, \( e^{-4t}B = 25e^{-4t} + C \).

Step 6: Solve for \( B \)

Multiply both sides by \( e^{4t} \) to solve for \( B(t) \):\[ B(t) = 25 + Ce^{4t} \]

Step 7: Apply the initial condition

Use the initial condition \( B(0) = 20 \) to find \( C \). Substitute \( t = 0 \) and \( B = 20 \) into the equation:\[ 20 = 25 + C \, e^{0} \]\[ 20 = 25 + C \]\[ C = -5 \]

Step 8: Write the particular solution

Substitute \( C = -5 \) back into the equation for \( B(t) \):\[ B(t) = 25 - 5e^{4t} \]

Key Concepts

First-Order Linear Differential Equation

A first-order linear differential equation is a type of equation that involves an unknown function and its derivative. It has the general form:

• \( \frac{dy}{dt} + P(t)y = Q(t) \)

In this equation, \( y \) is the function of \( t \) we want to find, and \( P(t) \) and \( Q(t) \) are given functions of \( t \). The term "first-order" indicates that the equation involves the first derivative of the unknown function.

For example, in the given problem, the differential equation \( \frac{dB}{dt} = 4B - 100 \) is identified as a first-order linear differential equation. It can also be rewritten as:

• \( \frac{dB}{dt} - 4B = -100 \)

Here, if we compare it to the general form, \( P(t) = -4 \) and \( Q(t) = -100 \). Understanding this structure is essential because it allows us to apply specific techniques to find solutions effectively.

Integrating Factor

The integrating factor is a powerful tool used to solve first-order linear differential equations. It helps transform the equation into a form that can be easily integrated. The integrating factor \( \mu(t) \) is given by:

• \( \mu(t) = e^{\int P(t) \, dt} \)

By multiplying the entire differential equation by this factor, the left side becomes a perfect derivative, simplifying the solving process.

In our case, where the equation is \( \frac{dB}{dt} - 4B = -100 \), we have \( P(t) = -4 \). So, the integrating factor becomes:

• \( \mu(t) = e^{-4t} \)

Multiplying the entire equation by \( e^{-4t} \), we simplify the problem:

• \( e^{-4t} \frac{dB}{dt} - 4B e^{-4t} = -100 e^{-4t} \)

This transformation allows the equation's left side to simplify to the derivative \( \frac{d}{dt}(e^{-4t}B) \), making integration straightforward.

Initial Condition

The initial condition in a differential equation provides a specific value for the unknown function at a particular point. This condition is crucial for finding a unique solution from a family of solutions.

In the example, the initial condition specifies that when \( t = 0 \), \( B = 20 \). Applying this condition helps us find the arbitrary constant \( C \) in the general solution.

• Substitute \( t = 0 \) and \( B = 20 \) into the equation \( B(t) = 25 + Ce^{4t} \)
• \( 20 = 25 + C \cdot 1 \)
• Solve for \( C \) to find \( C = -5 \)

By using the initial condition, we arrive at a particular solution:

• \( B(t) = 25 - 5e^{4t} \)

This solution satisfies both the differential equation and the initial condition, ensuring it is the unique solution corresponding to the problem presented.