91Ó°ÊÓ

The characteristic equation is a vital part of solving a linear differential equation with constant coefficients. To understand how it forms, we start with the differential equation: \[2y'' - 4y' + y = 0\].
In these types of problems, we assume a solution of the form \(y = e^{rx}\), where \(r\) is a constant.
By substituting \(y, y', \) and \(y''\) in the equation, we replace them with \(e^{rx}, re^{rx}, \) and \(r^2 e^{rx}\) respectively.

• This substitution transforms the differential equation into an algebraic equation in terms of \(r\).
• Therefore, in our given equation, after substitution and simplification, we get the characteristic equation \(2r^2 - 4r + 1 = 0\).

The roots of this equation will guide us in finding the general solution to the differential equation. It's a polynomial equation where the degree tells us the number of roots, and hence, the number of fundamental solutions to the differential equation.

The quadratic formula is a well-known tool for solving quadratic equations, which have the form \(ax^2 + bx + c = 0\). For our derived characteristic equation, \(2r^2 - 4r + 1 = 0\), we use the formula:
\[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
Here, \(a = 2\), \(b = -4\), and \(c = 1\).

• First, compute the discriminant, \(b^2 - 4ac\). For our equation, it becomes \((-4)^2 - 4 \cdot 2 \cdot 1 = 16 - 8 = 8\).
• Next, apply the quadratic formula to get roots: \(r = \frac{4 \pm \sqrt{8}}{4}\).
• Simplifying further gives: \(r = 1 \pm \frac{\sqrt{2}}{2}\).

This calculation shows two distinct real roots for our characteristic equation. These roots correspond to the exponents of \(e^x\) in the general solution of the differential equation.

The general solution of a differential equation is a combination of all possible solutions and usually contains constant coefficients that get determined based on initial conditions.
In the context of our problem, after obtaining the roots from the characteristic equation, \(r_1 = 1 + \frac{\sqrt{2}}{2}\) and \(r_2 = 1 - \frac{\sqrt{2}}{2}\), we form the general solution.

• Each root corresponds to an exponential solution of the form \(e^{rx}\). So, our specific solutions are \(e^{(1 + \frac{\sqrt{2}}{2})x}\) and \(e^{(1 - \frac{\sqrt{2}}{2})x}\).
• The general solution is a linear combination of these solutions: \(y(x) = C_1 e^{(1 + \frac{\sqrt{2}}{2})x} + C_2 e^{(1 - \frac{\sqrt{2}}{2})x}\).
• Where \(C_1\) and \(C_2\) are arbitrary constants that will be determined by any initial or boundary conditions presented with the differential equation.

This final expression represents the complete set of solutions for the differential equation under consideration.