Second-Order Homogeneous Linear ODE with Constant Coefficients

Definition

$y^{\prime \prime }=a{y}^{\prime }+by=0,a,b\in \mathbb{R}$

Solution

Assume the solutions will be $y=e^{\lambda x}$ form then ${\lambda }^2{e}^{\lambda x}+a\lambda e^{\lambda X}+b{e}^{\lambda x}=({\lambda }^2+a\lambda +b)\underset{>0}{\underbrace{e^{\lambda x}}}=0$ $\Rightarrow {\lambda }^2+a\lambda +b=0$^[characteristic equation]

${\lambda }_{1,2}=\frac{-a\pm \sqrt{a^2-4b}}{2}\Rightarrow y_1=e^{{\lambda }_{1}x}$, $y_2=e^{{\lambda }_{2}x}$ are solutions

Two Distinct Real Roots ${\lambda }_1\ne {\lambda }_2$

General Solution

$y=c_1{e}^{{\lambda }_{1}x}+c_2{e}^{{\lambda }_{2}x},c_1,c_2\in \mathbb{R}$

Multiple Real Roots $\lambda =-\frac{a}{2}$

$y_1=e^{\lambda x}=e^{-\frac{a}{2}x}$

by using the Reduction of Order, we can get $y_2$.

$U=\frac{1}{y_1^2}{e}^{-\int pdx}=\frac{1}{e^{-ax}}{e}^{-ax}=1$ $\Rightarrow y_2=u{y}_1=y_1\int Udx=e^{-\frac{a}{2}x}x=x{e}^{\lambda x}$

General Solution

$y=c_1{e}^{\lambda x}+c_{2}x{e}^{\lambda x}=(c_1+c_{2}x)e^{\lambda x},c_1,c_2\in \mathbb{R}$

Complex Roots ${\lambda }_1=-\frac{a}{2}+i\omega ,{\lambda }_2=-\frac{a}{2}-i\omega$

$\begin{array}{rl}\Rightarrow & {\hat{y}}_1=e^{-\frac{a}{2}x}{e}^{+i\omega x}=e^{-\frac{a}{2}x}(\cos \omega x+i\sin \omega x)\\ & {\hat{y}}_2=e^{-\frac{a}{2}x}{e}^{-i\omega x}=e^{-\frac{a}{2}x}(\cos \omega x-i\sin \omega x)\\ \Rightarrow & y_1=\frac{{\hat{y}}_1+{\hat{y}}_2}{2}=e^{-\frac{a}{2}x}\cos \omega x\\ & y_2=\frac{{\hat{y}}_1-{\hat{y}}_2}{2i}=e^{-\frac{a}{2}x}\sin \omega x\end{array}$

General Solution

$y=A{e}^{-\frac{a}{2}x}\cos \omega x+B{e}^{-\frac{a}{2}x}\sin \omega x,A,B\in \mathbb{R}$

Examples

$$\begin{array}{rl}& y^{\prime \prime }+5y^{\prime }+6y=0\\ & \text{assume}y=e^{\lambda x}\\ & {\lambda }^2+5\lambda +6=0\Rightarrow (\lambda +3)(\lambda +2)=0\Rightarrow \lambda =-3\text{or}-2\\ & y=c_1{e}^{-3x}+c_2{e}^{-2x}\end{array}$$