91Ó°ÊÓ

We start by substituting our guess, \(y = e^{rx}\), into the given equation. To do this, we first need to find the first and second derivatives of \(y\): \(y' = re^{rx}\), and \(y'' = r^2e^{rx}\). Now, substitute these into the given equation: \(r^2e^{rx} + pre^{rx} + qe^{rx} = 0\).

We can now factor out the term \(e^{rx}\) from the equation: \(e^{rx}(r^2 + pr + q) = 0\). Since \(e^{rx}\) is never zero, we can divide both sides of the equation by it: \(r^2 + pr + q = 0\).

The equation \(r^2 + pr + q = 0\) is called the characteristic equation of the given differential equation. This is a quadratic equation in \(r\) and can be solved using different methods, such as factoring, completing the square, or the quadratic formula: \(r = \frac{-p \pm \sqrt{p^2 - 4q}}{2}\). There are three main cases depending on the discriminant \(D = p^2 - 4q\): 1. \(D > 0\): The equation has two distinct real roots \(r_1\) and \(r_2\). The general solution is then \(y(x) = C_1e^{r_1x} + C_2e^{r_2x}\), where \(C_1\) and \(C_2\) are constants to be determined from the initial conditions. 2. \(D = 0\): The equation has one real root \(r_1 = r_2\). The general solution becomes \(y(x) = (C_1 + C_2x)e^{r_1x}\). 3. \(D < 0\): The equation has complex conjugate roots \(r_1 = \alpha + \beta i\) and \(r_2 = \alpha - \beta i\). The general solution can be written as \(y(x) = e^{\alpha x}(C_1\cos(\beta x) + C_2\sin(\beta x))\).

Depending on the specific case, as described above, we will end up with one of the three types of general solutions. If any initial conditions are given, we can use them to find the values of the constants \(C_1\) and \(C_2\). Otherwise, the result is the general solution to the given second-order linear differential equation.