91Ó°ÊÓ

In linear ordinary differential equations, especially those with constant coefficients, finding the *characteristic equation* is a crucial step for solving the differential equation. For a given second-order linear ODE like \( ay'' + by' + cy = 0 \), the characteristic equation is derived by replacing \( y \) with \( e^{rx} \) and its derivatives accordingly. This substitution simplifies to the algebraic equation \( ar^2 + br + c = 0 \). Finding the roots of the characteristic equation helps us understand the nature of the solutions to the differential equation. In our example, the characteristic equation is \( r^2 - 2r - 2 = 0 \). This results from identifying \( a = 1 \), \( b = -2 \), and \( c = -2 \) in the standard form.

Once we determine the roots of the characteristic equation, the next step is to construct the *general solution* of the differential equation. This depends heavily on the nature of the roots obtained from the characteristic equation.

• If the roots are distinct and real, like in our example, the general solution takes the form \( y = C_1 e^{r_1 x} + C_2 e^{r_2 x} \).
• If the roots are real and equal, then the general solution changes to \( y = C_1 e^{rx} + C_2 xe^{rx} \).
• Finally, if the roots are complex conjugates, the solution becomes \( y = e^{ ext{Re}(r)x} (C_1 \cos( ext{Im}(r) x) + C_2 \sin( ext{Im}(r) x)) \).

In this exercise, the distinct roots are \( r_1 = 1 + \sqrt{3} \) and \( r_2 = 1 - \sqrt{3} \), leading to the solution \( y = C_1 e^{(1+\sqrt{3})x} + C_2 e^{(1-\sqrt{3})x} \).

Differential equations with *constant coefficients* form an important class of equations that are easier to handle mathematically. In the given equation \( y'' - 2y' - 2y = 0 \), the constants are \( a = 1 \), \( b = -2 \), and \( c = -2 \), which appear as coefficients of the terms like \( y'', y', \) and \( y \) respectively. These coefficients do not vary with the independent variable \( x \), which makes solving these equations more straightforward. This invariant nature allows the powerful technique of using the characteristic equation to convert a differential equation problem into an algebra one, solvable with methods such as the quadratic formula.

To solve the characteristic equation of our example \( r^2 - 2r - 2 = 0 \), we turn to the *quadratic formula*. This formula provides the roots for any quadratic equation \( ar^2 + br + c = 0 \) and is given by \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). By substituting \( a = 1 \), \( b = -2 \), and \( c = -2 \) into the formula, we find:\[\begin{align*}r &= \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \ &= \frac{2 \pm \sqrt{4 + 8}}{2} \ &= \frac{2 \pm \sqrt{12}}{2} \ &= 1 \pm \sqrt{3}.\end{align*}\]The roots are \( r_1 = 1 + \sqrt{3} \) and \( r_2 = 1 - \sqrt{3} \), providing valuable insight into the behavior of the solutions to the original differential equation.