91Ó°ÊÓ

When dealing with differential equations, it's important to distinguish between homogeneous and non-homogeneous forms. In this context, a non-homogeneous differential equation refers to one that has a component independent of the function that we're solving for. In the exercise, we have \[ y'' + 2y' - 8y = 1 - 2x^2 \]. The term \(1 - 2x^2\) on the right-hand side is what makes this equation non-homogeneous. This means the solution will have to account for both the part that satisfies the homogeneous portion of the equation (where the right-hand side would be zero) and accommodate this additional term. Using the method of undetermined coefficients, we find a particular solution that helps in solving the non-homogeneous part.

The homogeneous solution deals with the equation obtained by setting the non-homogeneous term to zero. For the given exercise, this results in the following homogeneous differential equation: \[ y'' + 2y' - 8y = 0 \].To solve this, one typically finds a **characteristic equation**. This involves transforming the differential equation into an algebraic equation (commonly quadratic) often by substituting a potential exponential solution. The characteristic equation in this scenario is:\[ r^2 + 2r - 8 = 0 \].The solutions, or roots, to this equation will allow us to construct the general form of the homogeneous solution. For this exercise, the roots were found to be \( r = 2 \) and \( r = -4 \), leading to a homogeneous solution of the form:\[ y_h = C_1e^{2x} + C_2e^{-4x} \].This part of the solution addresses only the **structural** part of the equation without external forces or inputs.

The characteristic equation plays a fundamental role in solving linear homogeneous differential equations. It essentially translates the differential equation into a polynomial, the solutions of which give us information about the types of functions – typically exponential – that can satisfy the homogeneous equation. In our problem, by substituting a trial solution of the form \( y = e^{rx} \), differentiating it, and substituting into the homogeneous equation, we obtained the characteristic equation: \[ r^2 + 2r - 8 = 0 \].Solving this quadratic equation using standard methods such as factoring, completing the square, or the quadratic formula, we get real roots \( r = 2 \) and \( r = -4 \). These roots indicate that the solution to the homogeneous equation will be a combination of exponential functions: \[ y_h = C_1e^{2x} + C_2e^{-4x} \].Each root corresponds to a basis solution that fits the exponential form linked with the equation.

Matching coefficients is a crucial step when using the method of undetermined coefficients to find the particular solution. After choosing an appropriate form for the particular solution, based on the non-homogeneous term, you substitute this guess along with its derivatives into the non-homogeneous differential equation.For our exercise, the non-homogeneous term \(1 - 2x^2\) suggested trying a polynomial particular solution of the form: \[ y_p = Ax^2 + Bx + C \].Upon substituting into the differential equation and simplifying, you get \[(8A)x^2 + (4A + 2B)x + (2A + 2B - 8C) = 1 - 2x^2 \].To satisfy this equation, the coefficients of corresponding powers of \(x\) on both sides must be equal. This leads to a system of equations:

• \(8A = -2\)
• \(4A + 2B = 0\)
• \(2A + 2B - 8C = 1\)

Solving these gives the values \( A = -\frac{1}{4}, B = \frac{1}{2}, C = -\frac{1}{8} \). These coefficients define the particular solution we sought, complying with the polynomial structure of the non-homogeneous term.