91Ó°ÊÓ

The method of variation of parameters is a powerful technique used to find a particular solution to a non-homogeneous differential equation. This method involves using the solutions of the corresponding homogeneous equation to help construct a particular solution. In this exercise, once the complementary solution is determined, we assume a form for the particular solution:

• Assume: \( y_p = u_1(x)e^{5x} + u_2(x)e^{-x} \)

This assumption leverages the solutions from the homogeneous differential equation to find functions \( u_1(x) \) and \( u_2(x) \) that satisfy the given non-homogeneous equation.
By taking derivatives and substituting them back into the original differential equation, a system of equations is derived. These equations are structured so that one of the terms cancels out to simplify the process of finding \( u_1(x) \) and \( u_2(x) \). By integrating these functions, we establish the particular solution for the differential equation.

The method of undetermined coefficients is another approach used to find a particular solution to linear differential equations with constant coefficients. This technique is more straightforward than the variation of parameters but can only be applied in specific cases. Here, we form a trial solution based on the type of the non-homogeneous term.

• Try: \( y = Ae^x + B \)

We choose our trial solution based on the form of the non-homogeneous part of the equation \( e^x + 4 \). The strategy involves guessing a form for the particular solution and then substituting it into the differential equation.
By matching coefficients from the resulting equation, we can solve for the unknowns \( A \) and \( B \). This makes it possible to obtain the particular solution for the problem at hand. This method is particularly useful when the non-homogeneous term is a polynomial, exponential, sine, or cosine function.

In both methods mentioned, finding the complementary solution is an essential first step. This involves solving the related homogeneous differential equation:

• Homogeneous Equation: \( \frac{d^2 y}{d x^2} - 4 \frac{d y}{d x} - 5 y = 0 \)

To do this, we solve the characteristic equation derived from assuming a solution of the form \( y = e^{rx} \). For this exercise, the characteristic equation is:
\[ r^2 - 4r - 5 = 0 \]Solving this quadratic equation using the quadratic formula yields the roots \( r = 5 \) and \( r = -1 \). These roots provide the basis for the complementary solution:

• \( y_c = C_1 e^{5x} + C_2 e^{-x} \)

This solution reflects the behavior of the differential equation's inherent dynamics without external influences. It's crucial because it forms the backbone to which the particular solution is added to obtain the general solution.

The particular solution of a differential equation is the part that covers the specific influences or forcing functions in the equation. This solution, together with the complementary solution, forms the general solution of a differential equation.
In variation of parameters, \( y_p \) is constructed using functions derived from the integral of equations formed with respect to the non-homogeneous term. With undetermined coefficients, \( y_p \) is similarly constructed by assuming a functional form that mirrors the non-homogeneous function and solving for constants.

• Example: Assume \( y = Ae^x + B \) and use coefficients to solve.

The importance of the particular solution lies in its representation of how external influences shape the system’s response over time. It's directly dependent on \( e^x + 4 \) in the context of this exercise, showcasing how different methods can lead to the same functional outcome in principle.