Solving for the complementary function involves finding the general solution to the homogeneous version of a given differential equation. Here's a breakdown of the process:
1. Understand the Context:
- The complementary function (yc) is part of the general solution to a non-homogeneous differential equation. The general solution is the sum of the complementary function and the particular solution (yp): y = yc + yp.
- The particular solution (yp) addresses the non-homogeneous part of the differential equation (the part that doesn't equal zero). The complementary function focuses solely on the homogeneous equation.
2. Form the Homogeneous Equation:
- If your differential equation is in the form: ay'' + by' + cy = f(x), where a, b, and c are constants, the homogeneous equation is: ay'' + by' + cy = 0. Essentially, you set the non-zero right-hand side (f(x)) to zero.
3. Solve the Homogeneous Equation:
This typically involves finding the roots of the characteristic equation.
- Assume a solution of the form y = e^(rx), where 'r' is a constant.
- Substitute this assumed solution and its derivatives into the homogeneous differential equation.
- Simplify and factor out e^(rx). Since e^(rx) is never zero, the remaining polynomial must be zero. This polynomial is called the characteristic equation (also known as the auxiliary equation). For a second-order equation ay'' + by' + cy = 0, the characteristic equation is ar² + br + c = 0.
- Solve the characteristic equation for 'r'. The roots of this equation determine the form of the complementary function.
4. Determine the Form of the Complementary Function Based on the Roots:
The nature of the roots of the characteristic equation dictates the form of the complementary function. Here's a table summarizing the possibilities:
| Roots of Characteristic Equation (r) | Complementary Function (yc) | Example |
|---|---|---|
| Two distinct real roots (r1, r2) | yc = C1e^(r1x) + C2e^(r2x) | r1 = 2, r2 = 3 => yc = C1e^(2x) + C2e^(3x) |
| Repeated real root (r) | yc = (C1 + C2x)e^(rx) | r = 4 => yc = (C1 + C2x)e^(4x) |
| Complex conjugate roots (α ± βi) | yc = e^(αx) (C1cos(βx) + C2sin(βx)) | r = 1 ± 2i => yc = e^(x) (C1cos(2x) + C2sin(2x)) |
Where:
- C1 and C2 are arbitrary constants that are determined by initial conditions or boundary conditions.
- α is the real part of the complex root.
- β is the imaginary part of the complex root.
5. General Solution:
Once you've determined the complementary function (yc) and the particular solution (yp) (using methods appropriate for the non-homogeneous part of the equation), the general solution to the original differential equation is:
y = yc + yp
Example:
Solve the differential equation: y'' - 3y' + 2y = 0
- Homogeneous Equation: This is already in homogeneous form: y'' - 3y' + 2y = 0
- Characteristic Equation: r² - 3r + 2 = 0
- Solve for r: (r - 1)(r - 2) = 0 => r1 = 1, r2 = 2
- Complementary Function: Since we have two distinct real roots, the complementary function is yc = C1e^(x) + C2e^(2x)
- General Solution: (Since the original problem was the homogeneous equation): y = C1e^(x) + C2e^(2x)
In summary, solving complementary functions involves finding the solution to the homogeneous differential equation by assuming an exponential solution, deriving the characteristic equation, finding its roots, and then constructing the complementary function based on the nature of those roots. This complementary function then forms part of the general solution to the original (possibly non-homogeneous) differential equation.
