91Ó°ÊÓ

A linear non-homogeneous differential equation is a type of differential equation that can be expressed in the form \[ a_n(y^{(n)}) + a_{n-1}(y^{(n-1)}) + ext{...} + a_1(y') + a_0y = f(x) \] where \( a_n, a_{n-1}, ..., a_0 \) are constants and \( f(x) \) is a non-zero function of \( x \). In our exercise, the equation is \( D^2 y + y = \cos x \), where \( D^2 \) represents the second derivative. The term \( \cos x \) makes it non-homogeneous because it is not a zero function.
The principal features of such equations include:

• Order: determined by the highest derivative, here it is 2.
• Linear: because each term involving the function and its derivatives is linear in \( y \).
• Non-Homogeneous: due to the presence of the non-zero function \( \cos x \).

The solution typically involves finding both the solution to the associated homogeneous equation and a particular solution that fits the non-homogeneous part.

The homogeneous equation is obtained by replacing \( f(x) \) with 0, yielding \( D^2 y + y = 0 \). Solving it requires finding a general solution by looking for solutions of the form \( y = e^{rx} \), which leads us to the characteristic equation for easier handling. Here, the form is converted to \( r^2 + 1 = 0 \).
This equation is solved by recognizing that it results in complex roots \( r = i \) and \( r = -i \). These complex roots indicate that the general solution for the homogeneous equation involves trigonometric functions:

• \( y_h = C_1 \cos x + C_2 \sin x \)

Finding these particular solutions involves using Euler's formula to express complex exponentials in terms of sine and cosine. This foundation sets us up for finding a particular non-homogeneous solution.

The particular solution of a non-homogeneous differential equation is a specific solution that makes the original equation true. For the equation \( D^2 y + y = \cos x \), the particular solution must fit the form of the non-homogeneous term, initially guessed but adjusted as needed.
Our first assumption of \( y_p = A \cos x + B \sin x \) produced no solution, as it led to a contradiction. This occurs because the particular solution needed an adjustment due to its resonance with the homogeneous solution. Thus, we apply the method of **undetermined coefficients** again but let \( y_p = Ax \cos x + Bx \sin x \).
Substituting back into the equation and solving gives us values for \( A \) and \( B \), thus providing a particular solution:

• \( y_p = -\frac{x}{2} \sin x \)

This particular solution captures the non-homogeneity of the original equation and, when added to the homogeneous solution, yields the complete solution for the differential equation.