91Ó°ÊÓ

A second-order linear differential equation is one that involves the second derivative of a function. These equations generally take the form \( a y'' + b y' + c y = f(x) \), where \( y'' \) is the second derivative, \( y' \) is the first derivative, and \( y \) is the original function. The coefficients \( a \), \( b \), and \( c \) are constants, and \( f(x) \) is a given function. In our problem, we have the specific equation:

• \( 5y'' + y' = -6x \)

This equation represents a relationship between the function, its derivatives, and a specified external function, \(-6x\). Here, it's our job to find a function \( y \) that satisfies this relationship while also adhering to the given initial conditions \( y(0) = 0 \) and \( y'(0) = -10 \).

Solving the homogeneous part of a second-order differential equation involves setting the equation equal to zero. For \( 5y'' + y' = 0 \), we ignore the external function \( f(x) \), which in this case is \(-6x\). The task is to find a solution that naturally arises from the structure of the equation itself.

• Assume the solution has the form \( y = e^{rt} \).
• Substituting this into the differential equation leads to the characteristic equation.
• The solutions to the characteristic equation determine the homogeneous solution.

For our equation, this process leads to the characteristic equation \( 5r^2 + r = 0 \). Solving this gives the roots \( r_1 = 0 \) and \( r_2 = -\frac{1}{5} \). These roots help us construct the general homogeneous solution:

• \( y_h = C_1 + C_2 e^{-\frac{x}{5}} \)

where \( C_1 \) and \( C_2 \) are constants to be determined later.

Finding a particular solution for a differential equation involves determining a specific solution that works with the non-homogeneous term \( f(x) \), which is \(-6x\) in this case. A straightforward approach is to assume a form of the particular solution that mirrors the type of function present in \( f(x) \).For the equation \( 5y'' + y' = -6x \), set the particular solution as \( y_p = Ax + B \), where \( A \) and \( B \) are constants. By substituting this assumed solution into our original differential equation, we solve for these constants. In this instance:

• Derivatives are \( y_p' = A \) and \( y_p'' = 0 \).
• Substituting, we find \( A = -6 \), leading to the particular solution \( y_p = -6x \).

This solution addresses the influence of the non-homogeneous term, completing one part of our final solution.

The characteristic equation is a crucial tool when solving homogeneous differential equations. It is derived by substituting an assumed exponential solution like \( y = e^{rt} \) into the differential equation. This translates the problem into a polynomial equation in terms of \( r \), known as the characteristic equation.In the exercise, substituting into the homogeneous equation:

• \( 5y'' + y' = 0 \)
• transforming into the characteristic form \( 5r^2 + r = 0 \).

The solution of the characteristic equation, \( 5r^2 + r = 0 \), involves factorization:

• \( r(5r + 1) = 0 \)
• giving roots \( r_1 = 0 \) and \( r_2 = -\frac{1}{5} \).

These roots tell us the fundamental solutions of the homogeneous part, allowing us to write down the general solution as:

• \( y_h = C_1 + C_2 e^{-\frac{x}{5}} \)

Understanding the characteristic equation is essential as it forms the basis for constructing the general solution to differential equations with constant coefficients.