91Ó°ÊÓ

Second-order differential equations involve derivatives up to the second order, meaning they include terms like \( y'' \), which is the second derivative of the function \( y \) that you are trying to find. These types of differential equations are used to model many physical phenomena, such as the motion of a simple pendulum, the oscillations of a string, or electrical circuits.
An equation like \( D^2y - y = e^{-x} \) indicates that the second derivative \( D^2y \) and the function \( y \) are related, and you need to solve for the function \( y \). Often, these equations are linear, meaning the function and its derivatives appear to the power of one.
To solve second-order differential equations, a common approach is to first solve the homogeneous part (where the equation is set to zero) and then find a particular solution for the non-zero part. Combining these solutions gives us the general solution.

In a differential equation with constant coefficients, the coefficients of the derivatives are constants, not functions of \( x \). This makes solving the equation more straightforward, as it often reduces the problem to solving a polynomial equation.
For example, in the differential equation \( D^2y - y = e^{-x} \), the coefficients of \( D^2y \) and \( y \) are 1 and -1, respectively. Because these coefficients do not change, this is an equation with constant coefficients.
Such equations are typically solved using a method that involves finding the roots of the characteristic equation, which is derived from the differential equation itself. These roots help to form the complementary function, which is a key part of the general solution.

Finding a particular solution involves solving the non-homogeneous part of the differential equation. A particular solution satisfies the equation but does not include arbitrary constants, unlike the complementary function.
In our exercise, a hint is provided for the form of the particular solution: \( y_p = Ax e^{-x} \). This form is carefully chosen based on the non-homogeneous part of the equation \( e^{-x} \). By differentiating \( y_p \) and substituting back into the differential equation, we can solve for the constant \( A \).
Substituting gives an expression involving \( A \) which is matched to the known non-homogeneous term \( e^{-x} \). Solving for \( A \) completes the particular solution, which is then combined with the complementary function to form the general solution.

The homogeneous equation is a simplified version of the original differential equation where the right-hand side is set to zero. For the problem at hand, the homogeneous equation is \( D^2y - y = 0 \).
To solve this, we find the characteristic equation, \( \lambda^2 - 1 = 0 \), which is derived directly from the differential equation. The solutions to this characteristic equation, known as roots, help us form the complementary function. Here, the roots \( \lambda = 1 \) and \( \lambda = -1 \) result in the complementary function \( y_c = C_1e^x + C_2e^{-x} \).
The complementary function represents the general solution to the homogeneous part and contains constants \( C_1 \) and \( C_2 \) that are determined by initial conditions or additional information given in a specific problem.