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An integrating factor is a useful mathematical tool that simplifies solving linear first-order differential equations. Essentially, an integrating factor transforms a differential equation into one that is easily integrable.
To find the integrating factor, first express your linear differential equation in standard form which looks like this:

• \( \frac{dy}{dt} + P(t)y = Q(t) \)

For the given problem, \( P(t) = 2 \) and \( Q(t) = 1 \).
The integrating factor \( \mu(t) \) is obtained by taking the exponential of the integral of \( P(t) \):

• \( \mu(t) = e^{\int P(t) \, dt} = e^{2t} \)

This integrating factor, \( e^{2t} \), is then used to multiply every term in the equation. The goal is to rewrite the left side in a form that is the derivative of a product of two functions. This manipulation makes the equation easy to integrate as seen in the original solution plan.

An implicit solution represents a solution to the differential equation where terms involve the dependent variable (like \( y \)) in a non-isolated form. In other words, \( y \) isn't written just in terms of other known quantities.
For the equation \( \frac{dy}{dt} + 2y = 1 \), after multiplying by the integrating factor \( e^{2t} \), the left-hand side becomes a derivative:

• \( \frac{d}{dt}(e^{2t} y) = e^{2t} \)

Once integrated, this becomes:

• \( e^{2t} y = \frac{1}{2} e^{2t} + C \)

This is an implicit solution. Notice \( y \) is not explicitly expressed alone on one side of the equation; instead, it's still part of a relationship involving exponential functions. Implicit solutions are sometimes preferred, particularly when they make further analysis or computation more straightforward.

An explicit solution clearly expresses the dependent variable, \( y \), as a function of the independent variable, \( t \), without any ambiguity.
Take the implicit solution \( e^{2t} y = \frac{1}{2} e^{2t} + C \) and solve it for \( y \):

• \( y = \frac{1}{2} + Ce^{-2t} \)

This formulation is an explicit solution since \( y \) is isolated on one side of the equation.
To find the specific explicit solution for the initial-value problem, we use the initial condition \( y(0) = \frac{5}{2} \). By substituting this into the explicit form and solving for \( C \), we find:

• \( C = 2 \)

Thus, the explicit solution is \( y = \frac{1}{2} + 2e^{-2t} \). Explicit solutions are often more useful for understanding the behavior of the solution over time.

An initial-value problem is a type of differential equation along with a specified value for the unknown function at a given point, which sets the stage for finding a unique solution.
In this case, we begin with the equation:

• \( \frac{dy}{dt} + 2y = 1 \)

and the specified initial condition:

• \( y(0) = \frac{5}{2} \)

The initial condition provides a precise starting point to determine the constant \( C \) in the general solution. This condition distinguishes the particular solution from the family of possible solutions.
Using the initial condition in the explicit solution \( y = \frac{1}{2} + Ce^{-2t} \), we substituted \( t=0 \) and \( y=\frac{5}{2} \) to solve for \( C \). Finding \( C = 2 \) gives us the particular solution that meets the specific requirements of the problem. Initial-value problems are important in many scientific and engineering applications where initial conditions are known.