Why do we use $x^{m_1}$ instead of $e^{m_1x}$ for general solutions to Cauchy-Euler equations?

When I learned to calculate the general solution to homogeneous linear differential equations, I was told that for say,

$y''-y=0$, the auxiliary equation is $m^2-1=0\implies m^2=1\implies m=\pm 1$ and so the solution is $y=c_1e^{m_1x}+c_2e^{m_2 x}=c_1e^{x}+c_2e^{-x}$.

However, for Cauchy-Euler differential equations, which are linear and are homogeneous, the solutions take the form $y=c_1x^{m_1}+c_2x^{m_2}$.

This is the case for which the auxiliary equation provides two distinct real roots.

My question is if both the Cauchy-Euler equations and the non-Cauchy-Euler equations are homogeneous and they are both linear, why do one of the solutions involve $e^{m_1x}$ and the other involves just $x^{m_1}$? Why can't I just use $e^{m_ix}$ to find the solution even for the Cauchy-Euler ones?

• Because they are equidimensional equation. – IAmNoOne Mar 20 '14 at 19:57
• Short answer - those are different equations, so they have different set of solutions. There's a connection between Cauchy-Euler ODE and ODE with constant coefficients though, and you can easily see that connection if you do a substitution $x = e^u$. – Kaster Mar 20 '14 at 20:07

Short answer: they are actually the same solutions. It's just that $x$ that you see in the Cauchy-Euler equation corresponds to $e^x$ in the constant-coefficient equation. The Cauchy-Euler equation is the constant-coefficient equation in disguise. (The disguise being a change of independent variable.)
$$y'' + e^x y' + x y=0 \tag{1}$$ (which I just made up), those methods do not work, even if you try different substitutions for $y$ all day long. We simply cannot describe these solutions (other than $0$) in terms of functions for which we already have a name. Unless, of course, you create a new name for the solution - which is what people do (Bessel functions, Airy functions, etc). It seems nobody gave a name to the solutions of (1); you can be first.