Chapter 4.3, Problem 53P

If $x<0$, use the substitution $t=\ln \left(-x\right)$ to show that the general solutions of the Cauchy-Euler equation (i) in $x<0$ are identical to those in Problem $52$ with the exception that $x$ is replaced by $-x$.

Cauchy-Euler Equations. A (homegeneous) second order Cauchy-Euler Equation is one of the form

$a{x}^{2}y\text{'}\text{'}+bxy\text{'}+cy=0$, (i)

Where $a\ne 0$, $b,$ and $c$ are constants. If $x>0$, the substitution $t=\ln x$ and the chain rule for derivatives applied to $y\left(x\right)=y\left(t\left(x\right)\right)$,

$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{1}{x}\frac{dy}{dt}$ and $\frac{d^{2}y}{d{x}^2}=\frac{1}{x^2}\frac{d^{2}y}{d{t}^2}-\frac{1}{x^2}\frac{dy}{dt}$, transforms Eq. (i) into the constant coefficient linear equation

$a\frac{d^{2}y}{d{t}^2}+\left(b-a\right)\frac{dy}{dt}+cy=0$. (ii)

Thus, $y\left(t\right)$ is a solution of Eq. (ii), $y\ln \left(x\right)$ is a solution of Eq. (i) in $x>0$.

Show that the general solution of the Cauchy-Euler Equation (i) in $x>0$ is

$y=c_1{e}^{{\lambda }_{1}t}+c_2{e}^{{\lambda }_{2}t}=c_1{x}^{{\lambda }_1}+c_2{x}^{{\lambda }_2}$,

if ${\lambda }_1$ and ${\lambda }_2$ are real and ${\lambda }_1\ne {\lambda }_2$,

$y=c_1{e}^{{\lambda }_{1}t}+c_2{e}^{{\lambda }_{2}t}=c_1{x}^{{\lambda }_1}+c_2{x}^{{\lambda }_2}\ln x$,

if ${\lambda }_1={\lambda }_2$,

$y=e^{\mu t}\left[c_1\cos \left(vt\right)+c_2\sin \left(vt\right)\right]$

$=x^{\mu }\left[c_1\cos \left(v\ln x\right)+c_2\sin \left(v\ln x\right)\right]$,

if ${\lambda }_1=\mu +iv$ and ${\lambda }_2=\mu -iv$.