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A general solution to a differential equation is an expression that contains all possible solutions of the equation. It typically involves an arbitrary constant, which represents the family of solutions that can be achieved by varying its value. In the process of solving a first order linear differential equation, finding the general solution entails determining both the homogeneous and particular solutions.

To construct the general solution:

• First solve the homogeneous equation, which is obtained by setting the non-homogeneous term (the term without the derivative, in this case, the term involving \( t \)) to zero. This typically gives us a solution in terms of an arbitrary constant \( C \).
• Next, identify the particular solution using methods like variation of parameters, which is a technique that incorporates the effect of the non-homogeneous term.

The general solution itself is the sum of these homogeneous and particular solutions. It holds true under all circumstances, as it accounts for all possible initial conditions. For the equation given in the exercise, the general solution combines the homogeneous solution \( y_h = Ce^{-rac{1}{2} t} \) and the particular solution \( y_p = 2t - 4 \), hence \( y(t) = Ce^{-rac{1}{2}t} + 2t - 4 \).

An initial value problem (IVP) in differential equations is a problem where the solution is expected to satisfy certain initial conditions. Essentially, the IVP specifies the value of the unknown function at a given point, which allows us to determine the specific value of the arbitrary constant in the general solution.

In our problem, we are given the initial condition \( y(0) = 1 \). This means that when \( t = 0 \), the value of \( y \) is 1. By substituting this condition into the general solution, you can solve for the constant \( C \). Here’s how it works:

• Substitute \( t = 0 \) and \( y = 1 \) into the general solution \( y(t) = Ce^{-rac{1}{2}t} + 2t - 4 \).
• This yields \( C - 4 = 1 \), which simplifies to \( C = 5 \).

This process transforms the general solution into a particular solution that fits the initial condition, making it the unique solution to the IVP. For this exercise, substituting \( C = 5 \) into the general solution gives the particular solution \( y(t) = 5e^{-rac{1}{2}t} + 2t - 4 \).

First order linear differential equations are type of differential equations characterized by containing only the first derivative of the unknown function. The general form of such an equation is \( y' + p(t)y = g(t) \), where \( p(t) \) and \( g(t) \) are functions of \( t \).

To solve these equations, one effective method is the method of integrating factors. However, in this exercise, we employ the method of variation of parameters to find a particular solution. Here's a brief process on dealing with first order linear differential equations:

• Solve the associated homogeneous equation, \( y' + p(t)y = 0 \), by separating variables or through direct integration if possible.
• Obtain the homogeneous solution in the form \( y_h(t) = Ce^{P(t)} \), where \( P(t) \) is an integral of \( p(t) \).
• For the non-homogeneous part, apply variation of parameters by assuming a particular solution of the form \( y_p = v(t)e^{P(t)} \) and finding \( v(t) \).

For the given equation \( y' + \frac{1}{2}y = t \), the homogeneous solution was found to be \( Ce^{-rac{1}{2}t} \), and by variation of parameters, the particular solution \( y_p = 2t - 4 \) was obtained.

Integration by parts is a fundamental technique used to integrate products of functions. It is derived from the product rule for differentiation and is particularly useful in solving integrals that are impractical to solve by elementary integration techniques.

The formula for integration by parts is given by:\[\int u \cdot dv = uv - \int v \cdot du\]where \( u \) and \( dv \) are chosen from the integrand's components.

In our exercise, while solving for \( v(t) \), we encounter the need for integration by parts. Specifically, when integrating \( \int te^{\frac{1}{2}t} dt \), we choose:

• \( u = t \) resulting in \( du = dt \)
• \( dv = e^{\frac{1}{2}t} dt \) leading to \( v = 2e^{\frac{1}{2}t} \)

Using these, integration by parts gives us \( v(t) = 2te^{\frac{1}{2}t} - \int 2e^{\frac{1}{2}t} dt \). Completing the integration simplifies to \( v(t) = 2te^{\frac{1}{2}t} - 4e^{\frac{1}{2}t} + C \). This integral is part of finding the particular solution in the variation of parameters approach, showcasing how integration by parts can be applied in a broader problem-solving context.