91Ó°ÊÓ

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.

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.

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.