91Ó°ÊÓ

When tackling nonhomogeneous differential equations, verifying a particular solution is a crucial initial step. A particular solution is a specific function that satisfies the nonhomogeneous differential equation. For the given problem, the function provided as the particular solution is \( y_p = x - 1 \). To verify this, we need to confirm that this solution satisfies the differential equation when substituted into it.

The process involves the following steps:

• First, find the derivatives of \( y_p \). In this case, \( y_p' = 1 \) and \( y_p'' = 0 \).
• Next, substitute \( y_p \), \( y_p' \), and \( y_p'' \) back into the differential equation: \[ y'' - y' - 2y = 1 - 2x. \]
• Upon substitution, the equation simplifies and must equal the right-hand side of the equation \( 1 - 2x \).

In doing so, you confirm that \( y_p = x - 1 \) is indeed the particular solution since both sides match after substitution.

Finding the complementary solution is an essential step in resolving the overall differential equation. The complementary solution, usually denoted as \( y_c \), satisfies the homogeneous part of the differential equation.

For the equation \( y'' - y' - 2y = 0 \), we need to determine a solution where the right-hand side is zero. This is achieved by solving the characteristic equation, which is derived from the homogeneous equation without its nonhomogeneous terms.

The characteristic equation here is \( r^2 - r - 2 = 0 \). A quadratic equation, it can be factored into \((r - 2)(r + 1) = 0\), giving roots \( r = 2 \) and \( r = -1 \). These roots are used to form the complementary solution:

• \( y_c = C_1 e^{2x} + C_2 e^{-x} \)

This solution represents a general family of solutions fulfilling the homogeneous equation, crucial for assembling the final general solution.

The characteristic equation is a pivotal step in solving linear homogeneous differential equations. It provides a simple method for finding the solutions that form the complementary solution. In our exercise, the characteristic equation arises from setting the homogeneous part of the differential equation to zero: \( y'' - y' - 2y = 0 \).

To find it, we replace the derivatives in the equation with powers of \( r \), resulting in a polynomial equation: \( r^2 - r - 2 = 0 \). By factoring, \((r - 2)(r + 1) = 0\), we find the roots \( r = 2 \) and \( r = -1 \). These roots reveal the exponential terms that form the complementary solution:

• The corresponding solution is \( y = C_1 e^{2x} + C_2 e^{-x} \).

By setting the equation to zero, the characteristic equation ensures that the complementary solution can be formulated, a foundational step for determining the complete solution.

Initial conditions are vital for reducing the general solution of a differential equation to a specific, unique solution that fits particular requirements. They are values given for the solution and its derivatives at specific points, ensuring uniqueness in solutions. In our problem, the initial conditions are \( y(0) = 0 \) and \( y'(0) = 1 \).

The general solution of the equation is a combination of the complementary solution and the particular solution: \( y = (x - 1) + C_1 e^{2x} + C_2 e^{-x} \). We use the initial conditions to find specific values for the constants \( C_1 \) and \( C_2 \):

• Substitute \( x = 0 \) into \( y = 0 \) to get the first equation: \( -1 + C_1 + C_2 = 0 \), which simplifies to \( C_1 + C_2 = 1 \).
• Substitute \( x = 0 \) into \( y' = 1 \) for the derivative to get the second equation: \( 1 = 1 + 2C_1 - C_2 \), simplifying to \( 2C_1 - C_2 = 0 \).

Solving this system yields \( C_1 = \frac{1}{3} \) and \( C_2 = \frac{2}{3} \). With these values, the specific solution is \( y = (x - 1) + \frac{1}{3} e^{2x} + \frac{2}{3} e^{-x} \), satisfying both the differential equation and the initial conditions.