Second Order Differential Equation

Given a second order differential equation:

\begin{equation*} a y^{”} + b y^{’} + c y = 0 \end{equation*}

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

\begin{equation*} (ar^2 + br + c) e^{rx} = 0 \end{equation*}

where $e^{rx}\ne 0$. Than if we find proper $r$ which can make $a{r}^2+br+c=0$, the problem solved. And we call $a{r}^2+br+c=0$ 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 $b^2-4ac\le 0$, which means the characteristic equation has two complex roots $r\pm i\omega$, thus $y=e^{rx}[C_1\cos (\omega x)+C_2\sin (\omega x)]$.