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A non-homogeneous differential equation includes terms that do not depend on the function being solved for and its derivatives. An example can be seen in the given equation:

y'' + 4y = 8x² - 12x

The term on the right, 8x² - 12x, is the non-homogeneous part and does not depend on the function y or its derivatives. Because of this non-homogeneous term, the equation cannot be solved simply by using the methods for homogeneous equations. It involves finding particular and homogeneous solutions and combining them to get a general solution.

To solve a second-order linear differential equation, we first find the characteristic equation. Consider the homogeneous part of the given equation:

y'' + 4y = 0

We convert it into a characteristic equation by substituting y with a trial solution of the form e^(rt), leading to:

r² + 4 = 0

Solving for r gives us complex roots r = ±2i. These roots help us form the general solution of the homogeneous equation using sine and cosine functions:

y_h = C1 cos(2x) + C2 sin(2x)

Where C1 and C2 are arbitrary constants determined by initial conditions.

The particular solution of a non-homogeneous differential equation is a specific solution that fits the non-homogeneous part of the equation. In this context, we assume a solution y_p of the form:

y_p = Ax² + Bx + C

to match the polynomial on the right-hand side of our main equation. Derivatives of our assumed solution:

y_p' = 2Ax + B
y_p'' = 2A

Substituting these into the non-homogeneous equation and equating coefficients allows us to solve for the constants A, B, and C. After substitution and simplification:

4Ax² + 4Bx + (2A + 4C) = 8x² - 12x

we see that:
4A = 8 → A = 2
4B = -12 → B = -3
2A + 4C = 0 → 4 + 4C = 0 → C = -1

Thus, the particular solution is:

y_p = 2x² - 3x - 1.

The homogeneous solution to a differential equation is the response from solving the associated homogeneous equation, where the non-homogeneous term is set to zero. For our specific equation:

y'' + 4y = 0

We solve the characteristic equation for r, which provides us with roots r = ±2i. Using these roots, the homogeneous solution combines cosine and sine functions:

y_h = C1 cos(2x) + C2 sin(2x)

These combinations represent the complete response of the system without any external forcing terms (the non-homogeneous part). When combined with the particular solution, these form the general solution to the original differential equation:

y = y_h + y_pThis leads to our final answer:
y = C1 cos(2x) + C2 sin(2x) + 2x² - 3x - 1