91Ó°ÊÓ

When dealing with homogeneous linear differential equations, one key feature is the constant coefficients. This means the coefficients of all derivatives in the equation are constants, not variables. Let's take an example:
- Consider the differential equation: \(y'' + 3y' + 2y = 0\). Here, the coefficients 3 and 2 are constants.
Having constant coefficients simplifies the process of finding general solutions. It allows us to use methods like the characteristic equation to solve these equations more systematically. Such equations typically arise in physical phenomena like mechanics and electrical circuits, where proportional relationships between derivatives are governed by fixed factors.
In the exercise, the differential equation has constant coefficients derived from the characteristic polynomial, ensuring the form \(y'' + 20y' + 100y = 0\) matches the general solution.

The characteristic equation is crucial in solving linear differential equations with constant coefficients. It transforms a differential equation into an algebraic one, making it easier to solve. Here's the process:
- For a second-order differential equation like \(y'' + 20y' + 100y = 0\), assume a solution of the form \(y = e^{rx}\).
- Substitute \(y = e^{rx}\) into the differential equation.
- This substitution leads to an algebraic equation in \(r\), known as the characteristic equation.
In our exercise, we derived \((r + 10)^2 = 0\) as the characteristic equation. Solving this, we find \(r = -10\) with multiplicity 2. The roots of this characteristic equation reveal the behavior of solutions to the corresponding differential equation.

Repeated roots occur when the characteristic equation has roots with multiplicity greater than one. This means the solution requires extra terms for each repeat of the root to account for all conditions. In our situation, the characteristic equation \((r + 10)^2 = 0\) has \(r = -10\) as a repeated root with multiplicity 2.
For each repeated root, the associated general solution includes additional terms involving powers of \(x\). For a repeated root \(r\) with multiplicity, the terms are:

• \(c_1 e^{rx}\)
• \(c_2 x e^{rx}\)

In this exercise, the repeated root \(-10\) results in the solution \(y = c_1 e^{-10x} + c_2 x e^{-10x}\). This addition of \(x e^{-10x}\) allows for a complete expression satisfying all initial conditions or boundaries given the multiplicity of the root.