Euler-Cauchy Equations

Definition

$x^2{y}^{\prime \prime }+ax{y}^{\prime }+by=0(x\ne 0)$

Solution

Assume the solutions $y=x^m$ form
then $x^{2}m(m-1)x^{m-2}+axm{x}^{m-1}+b{x}^m=m(m-1)+am+b=0$

$\Rightarrow m^2+(a-1)m+b=0$^[auxiliary equation]

$m=\frac{(1-a)\pm \sqrt{(1-a{)}^2-4b}}{2}=\frac{1-a}{2}\pm \sqrt{\frac{(1-a{)}^2}{4}-b}$

Two Distinct Real Roots $m_1,m_2$

$y_1=x^{m_1},y_2=x^{m_2}$

General Solution

$y=c_1{x}^{m_1}+c_2{x}^{m_2},c_1,c_2\in \mathbb{R}$

Multiple Real Roots $m=\frac{1-a}{2}$

$y_1=x^m=x^{\frac{1-a}{2}}$

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

$U=\frac{1}{y_1^2}{e}^{-\int pdx}=\frac{1}{y_1^2}{e}^{-\int \frac{a}{x}dx}=\frac{1}{x^{1-a}}{e}^{-a\ln x}=\frac{1}{x^{1-a}}{x}^{-a}=\frac{1}{x},p=\frac{ax}{x^2}=\frac{a}{x}$ $\Rightarrow y_2=u{y}_1=y_1\int Udx=x^m\ln x(x>0)$

General Solution

$y=c_1{x}^m+c_2{x}^m\ln x,c_1,c_2\in \mathbb{R}$

Complex Roots $m=\mu \pm i\nu ,\mu ,\nu \in \mathbb{R}$

$\Rightarrow {\hat{y}}_1=x^{\mu +i\nu }=x^{\mu }{x}^{i\nu }=x^{\mu }(e^{\ln x}{)}^{i\nu }=x^{\mu }{e}^{i\nu \ln x}=x^{\mu }(\cos (\nu \ln x)+i\sin (\nu \ln x))$ ${\hat{y}}_2=x^{\mu -i\nu }=x^{\mu }{x}^{-i\nu }=x^{\mu }(e^{\ln x}{)}^{-i\nu }=x^{\mu }{e}^{-i\nu \ln x}=x^{\mu }(\cos (\nu \ln x)-i\sin (\nu \ln x))$ $\Rightarrow y_1=\frac{{\hat{y}}_1+{\hat{y}}_2}{2}=x^{\mu }\cos (\nu \ln x)$ $y_2=\frac{{\hat{y}}_1-{\hat{y}}_2}{2i}=x^{\mu }\sin (\nu \ln x)$

General Solution

$y=A{x}^{\mu }\cos (\nu \ln x)+B{x}^{\mu }\sin (\nu \ln x),A,B\in \mathbb{R}$

Examples

$$\begin{array}{rl}& x^2{y}^{\prime \prime }-3x{y}^{\prime }-12y=0\\ & \text{substitute}{x}^m:=y\\ & x^2(x^m{)}^{\prime \prime }-3x(x^m{)}^{\prime }-12(x^m)=0\\ & m(m-1)x^m-3m{x}^m-12x^m=0\\ & m^2-4m-12=0\Rightarrow (m-6)(m+2)=0\Rightarrow m=6\text{or}-2\\ & y=C_1{x}^6+C_2{x}^{-2}\end{array}$$