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An Ordinary Differential Equation (ODE) is an equation that involves a function and its derivatives. ODEs are pivotal in describing various phenomena such as physical, biological, and economic processes. They are typically represented with the notation \( \frac{dy}{dt} \), where \( y \) is the dependent variable and \( t \) is the independent variable. In our problem, we have the equation \( y^{\prime}(t) - 2y = 8 \), a first-order ODE because it includes the first derivative of \( y \).Solving ODEs helps us find the function \( y(t) \) that satisfies the equation over a given interval. Such problems are often more manageable when paired with initial conditions, which we'll discuss shortly. The focus is usually on finding a specific solution rather than a general one when initial conditions are present.

The integrating factor is a powerful technique for solving linear first-order ODEs. In our specific equation \( \frac{dy}{dt} - 2y = 8 \), it is a practical tool to transform a difficult problem into an easier one. The idea is to multiply the entire differential equation by a strategically chosen function, called the integrating factor, to make the left side of the equation a product derivative.For any linear ODE in the form \( \frac{dy}{dt} + P(t)y = Q(t) \), the integrating factor \( \mu(t) \) is determined by the exponential function involving the integral of \( P(t) \). So, for our equation, where \( P(t) = -2 \), we calculate \( \mu(t) \) as:

• \( \mu(t) = e^{\int -2dt} = e^{-2t} \).

After multiplying the entire original ODE by this integrating factor, we convert it into a form where the left side is a straightforward derivative, easing the path to integration.

Initial conditions are specific values assigned to the function and its derivatives at the starting point of an interval, providing a unique solution to a differential equation. In our problem, the initial condition given is \( y(0) = 0 \). When combined with the ODE, these conditions help pinpoint the specific solution from a family of possible solutions.Applying initial conditions usually occurs after you have found the general solution. Once you arrive at an expression for \( y(t) \), in this case, \( y(t) = -4 + Ce^{2t} \), you plug in \( t = 0 \) and \( y(0) = 0 \) to determine the constant \( C \). It works as follows:

• \( 0 = -4 + C \Rightarrow C = 4 \).

Thus, these conditions ensure that the solution accurately reflects the behaviour stipulated by the initial state.

Integrating the coefficient is an essential step when using the integrating factor approach for solving an ODE. This integral forms the basis of the integrating factor itself. Let's break it down in the context of the given ODE \( \frac{dy}{dt} - 2y = 8 \).Here, the coefficient of \( y \) is \(-2 \), and we need to integrate this with respect to \( t \) to find the integrating factor:

• \( \int -2 \, dt = -2t \).

The function \( e^{-2t} \) results from exponentiating the integral of the coefficient, guided by the formula \( \mu(t) = e^{\int{-2dt}} \). This operation is pivotal as it brings the derivative of a product, simplifying the ODE to:

• \( \frac{d}{dt}(ye^{-2t}) = 8e^{-2t} \).

Thus, integrating the coefficient is essential to forming an equation that is simple to solve by direct integration.