91Ó°ÊÓ

An integrating factor is a crucial tool used to solve first-order linear differential equations. It helps to transform the equation into a form that is easier to work with and solve. In our exercise, the differential equation is given as \( \frac{dy}{dt} + y = 2 \). We recognize this as a standard form for a first-order linear ordinary differential equation.
The general form can be written as:

• \( \frac{dy}{dt} + P(t)y = Q(t) \)

To solve it, we calculate the integrating factor \( \mu(t) \) using the expression:

• \( \mu(t) = e^{\int P(t) \, dt} \)

In our case, \( P(t) = 1 \), so:

• \( \mu(t) = e^{\int 1 \, dt} = e^t \)

This integrating factor \( e^t \) is then used to multiply through the entire equation, turning it into something more workable, which allows us to later identify it as the derivative of something else.

An initial-value problem in differential equations involves finding a specific solution that satisfies a given initial condition as part of the solution process. For our exercise, this initial condition is provided as \( y(0) = 1 \). This means that at time \( t=0 \), the value of \( y \) must equal 1.
Let's outline why this is important:

• Initial conditions give us the ability to find a particular solution among many potential solutions.
• They enable us to pinpoint the exact path or curve the solution follows.

By applying the initial condition at the very end of our solution steps, we are able to solve for the constant \( C \) in our general solution. This transforms the general solution \( y = 2 + Ce^{-t} \) into a unique, specific solution \( y = 2 - e^{-t} \) by setting \( C = -1 \).

A particular solution of a differential equation is one specific solution that satisfies the initial condition, as we discussed earlier with the initial-value problem. Finding a specific solution often involves determining unknown constants after solving the general form of the equation.
Here's how this plays out:

• After using the integrating factor and integrating, we end up with a general solution: \( y = 2 + Ce^{-t} \).
• The next step involves applying the initial condition \( y(0) = 1 \).
• This allows us to find the value of \( C \): in this case, \( C = -1 \).

Substituting \( C = -1 \) back into the general solution gives us the particular solution \( y = 2 - e^{-t} \). This solution is unique as it precisely fits both the differential equation and the initial condition, completing the problem's requirements.