Solving Cauchy-Euler Equation with Constant Coefficients

[find_the_fun]

I'm trying to solve \(\displaystyle a(x-x_0)y''+b(x-x_0)y'+cy-c=0\)
So I let \(\displaystyle y=(x-x_0)^m\) then \(\displaystyle y'=m(x-x_0)^{m-1}\) and \(\displaystyle y''=m(m-1)(x-x_0)^{m-2}\)
plugging in gives \(\displaystyle a(x-x_0)m(m-1)(x-x_0)^{m-2}+b(x-x_0)m(x-x_0)^{m-1}+c((x-x_0)^m-1)=0\)

now I want to find the values of m that make the equation 0, but factoring seems to be an impossible task?

 

[MarkFL]

I assume we are given the Cauchy-Euler equation:

\(\displaystyle a\left(x-x_0\right)^2y''+b\left(x-x_0\right)y'+cy=c\tag{1}\)

where \(\displaystyle 0<x-x_0\)

Making the substitution:

\(\displaystyle x-x_0=e^t\)

will transform the ODE into an equation with constant coefficients.

It follows from the chain rule that:

\(\displaystyle \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=\frac{dy}{dx}e^t=\left(x-x_0\right)\frac{dy}{dx}\)

and hence:

\(\displaystyle \left(x-x_0\right)\frac{dy}{dx}=\frac{dy}{dt}\tag{2}\)

Differentiating this with respect to $t$, we find from the product rule that:

\(\displaystyle \frac{d^2y}{dt^2}=\frac{d}{dt}\left(\left(x-x_0\right)\frac{dy}{dx}\right)=\left(x-x_0\right)\frac{d}{dt}\left(\frac{dy}{dx}\right)+\frac{dx}{dt}\frac{dy}{dx}=\)

\(\displaystyle \left(x-x_0\right)\frac{d^2y}{dx^2}\frac{dx}{dt}+\frac{dy}{dt}=\left(x-x_0\right)^2\frac{d^2y}{dx^2}+\frac{dy}{dt}\)

and hence:

\(\displaystyle \left(x-x_0\right)^2\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}-\frac{dy}{dt}\tag{3}\)

Substituting into (1), the expressions in (2) and (3), we obtain:

\(\displaystyle a\left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)+b\frac{dy}{dt}+cy=c\)

Collecting like terms, we obtain:

\(\displaystyle a\frac{d^2y}{dt^2}+(b-a)\frac{dy}{dt}+cy=c\tag{4}\)

From here, I would determine the homogeneous solution, and a particular solution via the method of undetermined coefficients, from which you will obtain the general solution for $y(t)$, and then back-substitute for $t$ to get the general solution $y(x)$ to the original ODE.