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Differential equations are mathematical equations that relate a function with its derivatives. They arise in various fields of science and engineering. For example, they are used to model population growth, heat conduction, and motion dynamics. A simple example is the first-order differential equation \( dy/dx = y \), which describes exponential growth or decay. The goal when working with differential equations is to find the unknown function that satisfies the equation. In our exercise, we are dealing with a second-order differential equation:

• It's called second-order because the highest derivative is of order two (\( y'' \)).
• The equation in question is \( y'' + 2y' + y = x^2 \), a type of linear differential equation as the unknown function and its derivatives appear linearly.

Understanding the structure of differential equations is key when learning methods to solve them, such as the method of undetermined coefficients.

The characteristic equation is a crucial tool for solving linear homogeneous differential equations. It helps determine the solutions for these equations by transforming the differential equation into an algebraic equation. Here's how it works:- Consider the homogeneous equation from our original problem: \( y'' + 2y' + y = 0 \).- By assuming a solution of the form \( y = e^{mx} \), and substituting into the differential equation, we derive the characteristic equation: \( m^2 + 2m + 1 = 0 \).- The roots of the characteristic equation directly give us the types of solutions for the differential equation.For our step-by-step solution:- The characteristic equation \( (m + 1)^2 = 0 \) indicates \( m = -1 \) is a repeated root.- Repeated roots imply that the solution for the homogeneous equation will include terms like \( C_1 e^{-x} \) and \( C_2 x e^{-x} \).The characteristic equation is a quick and efficient pathway to understanding the nature of solutions for linear differential equations.

A homogeneous differential equation is one in which all terms are dependent on the unknown function or its derivatives. The right-hand side of such an equation is zero. For example, the homogeneous version of our equation is:- \( y'' + 2y' + y = 0 \).The general approach to solving these equations involves:- Finding the characteristic equation as described above.- Solving for its roots to determine the general form of the solution.In our specific exercise, we found that:- The repeated root \( m = -1 \) gave rise to the general solution \( y_h = C_1 e^{-x} + C_2 x e^{-x} \).This solution provides the basis upon which the particular solution is added when addressing non-homogeneous equations, leading us to the complete general solution.

The particular solution addresses the non-homogeneous part of the differential equation. In our example, this is the term \( x^2 \). Here's how we find it:- Make an educated guess about its form—given \( x^2 \), we assume \( y_p = Ax^2 + Bx + C \).- Derive the first (\( y_p' = 2Ax + B \)) and second (\( y_p'' = 2A \)) derivatives.- Substitute these derivatives back into the left side of the differential equation to match it to \( x^2 \).- Equate coefficients from both sides to solve for \( A \), \( B \), and \( C \).After solving, we found:- \( A = 1 \), \( B = -4 \), \( C = 6 \)- The particular solution is \( y_p = x^2 - 4x + 6 \).This solution is crucial as it captures the non-homogeneous aspect of the original equation and combines with the homogeneous solution to form the general solution.

The general solution to a differential equation is the sum of the homogeneous and particular solutions. It encompasses all possible solutions to the differential equation. Let's summarize the process:1. Compute the solution to the homogeneous equation, here: \( y_h = C_1 e^{-x} + C_2 x e^{-x} \).2. Find the particular solution to the non-homogeneous equation, here: \( y_p = x^2 - 4x + 6 \).3. Add together both solutions to obtain the general solution: \[ y = C_1 e^{-x} + C_2 x e^{-x} + x^2 - 4x + 6 \] This general solution represents any function that satisfies the original differential equation. The constants \( C_1 \) and \( C_2 \) are determined by initial conditions or boundary values if provided. This step ensures our solution accounts for all possible scenarios described by the differential equation.