Depending on the solution of the auxiliary equation, the solution of the ODE is different: \(\displaystyle m=\frac{-b\pm\Delta}{2a}, \Delta\equiv b^{2}-4ac\)
1. \(\Delta>0\Rightarrow\) 2 real roots: \(m_{1},m_{2}\), both \(y_{1}=e^{m_{1}x},y_{2}=e^{m_{2}x}\) works. So the general solution is \(y_{general}=Ae^{m_{1}x}+Be^{m_{2}x}\)
- Alternative form: Because of the Euler's Equation in \(\mathbb{C}\), \(y(x)=e^{px}[\tilde{A}\cosh(qx)+\tilde{B}\sinh(qx)]\), where \(p,q\) are \(\displaystyle m=\underbrace{- \frac{b}{2a}}_p\pm \underbrace{\frac{\sqrt{\Delta}}{2a}}_{q}\).
- Why we need this form? When the initial condition includes \(y(0)\), the alt form would be much easier
2. \(\Delta<0\Rightarrow\) 2 complex roots: \(m_{1},m_{2}\in\mathbb{C}\), so the general solution is \(y(x)=Ce^{m_{1}x}+De^{m_{2}x}\)
- Alternative form: Because of the Euler's Equation, \(y(x)=e^{m_{r}x}[\tilde{C}\cos(m_{i}x)+\tilde{D}\sin(m_{i}x)]\), where \(m_{1,2}=m_{r}\pm im_{i}\).
3. \(\Delta=0\Rightarrow\) one \(m\) only! so the general solution is \(y(x)=Ae^{mx}+Bxe^{mx}=(A+Bx)e^{mx}\)