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A non-homogeneous linear differential equation is a type of equation that equates a linear differential operator applied to an unknown function with a known function. In simpler terms, it's an equation where:

• There are derivatives of an unknown function (like velocity or acceleration in physics).
• The equation is not equal to zero; instead, it is equal to some other function or constant.

Here's the form this type of equation usually takes:\[ay'' + by' + cy = g(x)\]Where:

• \(a, b,\) and \(c\) are constants.
• \(y\) is the unknown function.
• \(g(x)\) is a non-zero function that makes the equation non-homogeneous.

In our original exercise, the equation is already formatted as such:\[2y'' - 5y' - 12y = 6\]This tells us that instead of aiming for a solution equal to zero, we will find a solution that also accounts for the external influence represented by \(6\). This distinction is key because it implies that solutions to such equations must account for additional terms which change the behavior fundamentally from a homogeneous one.

The characteristic equation is fundamental in solving linear differential equations, especially when analyzing the homogeneous part. It transforms a differential equation into an algebraic one. This step involves assuming a solution of an exponential form, \(y = e^{rx}\), which simplifies differentiation.

For the homogeneous differential equation from the exercise:\[2y'' - 5y' - 12y = 0\] We substitute \(y = e^{rx}\) and its derivatives into the equation, leading to the characteristic equation:\[2r^2 - 5r - 12 = 0\] This quadratic can be solved using the quadratic formula:\[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] For our equation, \(a = 2\), \(b = -5\), and \(c = -12\). By calculating the discriminant and solving for \(r\), we find:\[r_1 = 4, \quad r_2 = -\frac{3}{2}\] These roots are critical because they determine the form of the general solution for the homogeneous equation, which is a combination of terms like \(C_1 e^{r_1 x} + C_2 e^{r_2 x}\).

Building the general solution for a non-homogeneous differential equation involves combining two parts:

• The solution to the homogeneous equation (often expressed in exponential terms derived from the characteristic roots).
• A particular solution which satisfies the entire original equation, including \(g(x)\).

In the exercise's context, the homogeneous solution, derived from the roots of the characteristic equation \(r_1 = 4\) and \(r_2 = -\frac{3}{2}\), is:\[y_h = C_1 e^{4x} + C_2 e^{-\frac{3}{2}x}\] Next, the particular solution \(y_p\) accounts specifically for the non-homogeneity of the equation. In this case, since \(6\) is a constant term, \(y_p\) can be assumed as a constant. This gives:\[y_p = -\frac{1}{2}\] The general solution is then the sum:\[y = y_h + y_p = C_1 e^{4x} + C_2 e^{-\frac{3}{2}x} - \frac{1}{2}\] By combining these components, we accommodate both the inherent behavior (given by the homogeneous solution) and the external factor(s) expressed by \(g(x)\). Understanding the construction of the general solution is crucial as it encompasses the total response of the system to not only its own past state but also external inputs.