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The homogeneous solution is the key to solving the complementary part of a non-homogeneous differential equation. In this case, we deal with a second-order linear differential equation. The homogeneous form is where the right side is zero: \[ y'' + y' - 6y = 0 \] To find the homogeneous solution, we assume a solution of the form \( y_h = e^{rx} \). This exponential function is plugged into the homogeneous equation. Doing so allows us to derive what is known as the characteristic equation. For this problem, the characteristic equation is a quadratic: \[ r^2 + r - 6 = 0 \] Solving this quadratic equation requires finding its roots. The roots tell us the exponents in the homogeneous solution. Factoring gives: \[ (r+3)(r-2) = 0 \] Thus, the roots are \( r_1 = -3 \) and \( r_2 = 2 \). The homogeneous solution, formed by these roots, is a linear combination of the individual solutions based on these roots:\[ y_h = C_1 e^{-3x} + C_2 e^{2x} \] Here, \( C_1 \) and \( C_2 \) are constants determined by initial conditions.

The goal of finding a particular solution is to solve the non-homogeneous part of the differential equation. For the given equation:\[ y'' + y' - 6y = 2x \] We look for a specific solution that satisfies this equation. The method of undetermined coefficients is effective here, especially when the non-homogeneous term is a polynomial, exponential, or sine/cosine function.We begin by assuming a form for the particular solution. In our case, since the right-hand side of the equation is \( 2x \), we use a polynomial form:\[ y_p = Ax + B \] Next, we take the derivatives:- \( y_p' = A \)- \( y_p'' = 0 \)Substituting \( y_p \), \( y_p' \), and \( y_p'' \) into the non-homogeneous equation, we simplify and collect like terms:\[ 0 + A - 6(Ax + B) = 2x \] This results in a system of equations after equating coefficients:- \(-6A = 2\)- \(A - 6B = 0\)By solving these equations, we find:- \( A = -\frac{1}{3} \) - \( B = -\frac{1}{18} \) Thus, our particular solution is:\[ y_p = -\frac{1}{3}x - \frac{1}{18} \]

The characteristic equation emerges from solving the homogeneous part of a linear differential equation. It is essential as it helps determine the form of the homogeneous solution. The process starts by substituting an assumed solution, like \( y = e^{rx} \), into the homogeneous version of the differential equation:\[ y'' + y' - 6y = 0 \] The assumption leads us to replace the derivatives with powers of \( r \) (because when \( y = e^{rx} \), derivatives are simply multiples of \( e^{rx} \)): - First derivative: \( ry_h = re^{rx} \)- Second derivative: \( r^2y_h = r^2e^{rx} \)Substituting these back, we get the characteristic equation:\[ r^2 + r - 6 = 0 \] Solving this quadratic equation by factoring, we find:\[ (r+3)(r-2) = 0 \]The solutions, \( r_1 = -3 \) and \( r_2 = 2 \), represent the parts of the homogeneous solution. This means:\[ y_h = C_1 e^{-3x} + C_2 e^{2x} \] These roots and the resulting homogeneous solution are crucial for building the overall solution to a differential equation.