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A homogeneous differential equation is a type of differential equation where every term is a function of the dependent variable and its derivatives. In simpler words, it doesn't have a standalone term that doesn't multiply the dependent variable or its derivatives. For example, the homogeneous equation in the problem, which is \( y''' - 9y'' + 27y' - 27y = 0 \), has each term involving a derivative of \( y \).
This equation is derived by setting the non-homogeneous part, which in this exercise is the external forcing function \( 54 \sin(3x) \), to zero.
To solve this equation, we assume a solution form involving an exponential function, usually \( y = e^{rx} \). This leads us to form a characteristic equation by plugging the assumed solution into the homogeneous equation. The goal is to find the values of \( r \) that satisfy the characteristic equation. These values are crucial because they guide us to the general solution of the homogeneous equation. In our problem, the characteristic equation is \( r^3 - 9r^2 + 27r - 27 = 0 \). By factoring out roots, we find that \( r = 3 \) repeats three times, hence the general solution is \( y_h = (C_1 + C_2 x + C_3 x^2) e^{3x} \).
This solution, formed by combining terms with \( C_1 \), \( C_2 \), and \( C_3 \), accounts for the multiplicity of the root.

A non-homogeneous solution is one approach to solve differential equations that have additional terms or functions that cannot be eliminated through simplification. These additional terms, known as forcing functions, necessitate a different method compared to a homogeneous differential equation. In the problem, this function is \( 54 \sin(3x) \).
To tackle the non-homogeneous part of the equation, we use a method called the "method of undetermined coefficients." This technique involves guessing a particular solution form that is similar to the forcing function. Here, we assume \( y_p(x) = A \cos(3x) + B \sin(3x) \), and substitute these guessed functions back into the differential equation to find the coefficients A and B that satisfy the equation.
Solving involves matching coefficients with like terms. For this problem, we discover that \( A = -\frac{3}{2} \) and \( B = 0 \). Hence, the particular solution for the non-homogeneous equation becomes \( y_p(x) = -\frac{3}{2} \cos(3x) \).
By combining this with the homogeneous solution, we form the general solution for the entire differential equation.

An initial value problem is a sophisticated phrase used to describe differential equations that come with specified values for the function and its derivatives at a certain point—usually at \( x = 0 \). This problem structure helps in discovering particular solutions from a general solution, tailoring it to specific conditions. In our exercise, the initial values are \( y(0) = 3.5 \), \( y'(0) = 13.5 \), and \( y''(0) = 38.5 \).
Once we obtain the general solution \( y(x) = y_h(x) + y_p(x) \), we can use these initial conditions to solve for the constants \( C_1, C_2, \) and \( C_3 \) included in the homogeneous solution part. Substituting these initial conditions into the general solution and its respective derivatives allows us to generate a system of equations that can be solved.

• At \( x=0 \), extract equations by substituting into \( y, y', y'' \).
• Solve the system of equations simultaneously to determine \( C_1, C_2, \) and \( C_3 \).

Solving these provides the specific expressions needed to determine the constants, particularly yielding \( C_1 = 5 \), \( C_2 = -\frac{9}{2} \), and \( C_3 = \frac{5}{4} \).
The final solution incorporates these constants, resulting in a unique solution that satisfies both the differential equation and the initial conditions.