Second Order Differential Equation

Given a second order differential equation:

$ a y^{''} + b y^{'} + c y = 0 $

Notice that the nth order differential of \(y = e^{rx}\) is \(r^n e^{rx}\), we can use this as the basic formation of \(y\),

$ (ar^2 + br + c) e^{rx} = 0 $

where \(e^{rx} \neq 0\). Than if we find proper \(r\) which can make \(ar^2 + br + c = 0\), the problem solved. And we call \(ar^2 + br + c = 0\) as Characteristic equation.

1. If \(b^2 - 4ac > 0\), which means the Characteristic equation has two different roots \(r_1, r_2\), thus \(y = C_1 e^{r_1 x} + C_2 e^{r_2 x}\)
2. If \(b^2 - 4ac = 0\), which means the Characteristic equation has two equal roots \(r_1\), thus \(y = C_1 e^{r_1 x} + C_2 x e^{r_2 x}\)
3. If \(b^2 - 4ac \leq 0\), which means the characteristic equation has two complex roots \(r \pm i\omega\), thus \(y = e^{rx} [C_1 \cos(\omega x) + C_2 \sin(\omega x)]\).