91Ó°ÊÓ

The complementary solution is a crucial component of the general solution of a differential equation. It represents the solution to the corresponding homogeneous equation, which, in this case, is given by \((D^3 + D^2 - 4D - 4)y = 0\). To find this solution, you need to determine the roots of the characteristic equation. For our example, the characteristic equation is \(r^3 + r^2 - 4r - 4 = 0\). These roots can be found using methods like the rational root theorem or synthetic division. In our exercise, the roots are \(r = 1\), \(r = -2\), and \(r = -2\), which results in a complementary solution of:

• \(y_c = C_1e^{x} + C_2e^{-2x} + C_3xe^{-2x}\)

Here, the constants \(C_1\), \(C_2\), and \(C_3\) are arbitrary and represent the freedom in the solution consistent with the initial conditions of the problem. Notice the presence of \(xe^{-2x}\) due to the repeated root \(-2\). This is a distinguishing aspect when dealing with repeated roots.

The particular solution is the other part of the general solution of a non-homogeneous differential equation. It's specifically constructed to satisfy the non-homogeneous part of the equation, which includes any terms not influenced by the complementary solution, such as polynomial or exponential functions. For the differential equation \((D^3 + D^2 - 4D - 4)y = 8x + 8 + 6e^{-x}\), the form of the particular solution is chosen based on the right-hand side prompts. Here, the terms \(8x + 8 + 6e^{-x}\) guide us to try a specific form:

• \(y_p = Ax + B + Ce^{-x}\)

We find this form by guessing the types of terms present in the non-homogeneous part. Substitute this form into the differential equation and solve for the coefficients \(A\), \(B\), and \(C\). This leads us to find that \(A = 2\), \(B = 6\), \(C = 2\). Thus, the particular solution here is:

• \(y_p = 2x + 6 + 2e^{-x}\)

Each term in the particular solution corresponds to a part of the non-homogeneous term, ensuring the equation balances properly when plugged back.

The characteristic equation is a vital step in solving linear differential equations with constant coefficients. It stems from the homogeneous part of the differential equation by assuming a solution of the form \(y = e^{rx}\). By substituting this assumed solution into the differential equation \((D^3 + D^2 - 4D - 4)y = 0\) and replacing each derivative with \(r\), we derive the characteristic equation:

• \(r^3 + r^2 - 4r - 4 = 0\)

The roots of this polynomial equation direct us to the form of the complementary solution. In our example, solving \(r^3 + r^2 - 4r - 4 = 0\) gives roots \(r = 1\), \(r = -2\), and \(r = -2\), leading us to the form \(C_1e^{x} + C_2e^{-2x} + C_3xe^{-2x}\). This is a pattern you'll see frequently: each distinct root contributes an exponential term, and repeated roots contribute multiple terms of increasing polynomial degree.When solving such equations, consider:

• The type and number of roots: real and distinct, repeated, or complex.
• The form of each resulting term in the complementary solution.

Completing this step correctly paves the way for understanding the full dynamics described by the differential equation.