the trial solution y = x^r
For a constant-coefficient equation the lucky guess was y = e^(rx), because exponentials reproduce themselves under differentiation. For a Cauchy-Euler equation the right lucky guess is different: y = x^r, a plain power. The reason is a perfect matchmaking — the equation's x-powers are arranged to cancel exactly the x-powers that differentiating x^r produces.
Watch the cancellation. If y = x^r then y' = r x^(r-1) and y'' = r(r-1) x^(r-2). In the Cauchy-Euler equation a x^2 y'' + b x y' + c y, the x^2 multiplies y'' to give a r(r-1) x^r, the x multiplies y' to give b r x^r, and c y is c x^r. Every term is now a number times the SAME x^r, so factoring it out leaves the indicial equation a r(r-1) + b r + c = 0 — a quadratic (degree-n in general) in r with no x left. Solve it for r. Each real root r yields a solution x^r; a repeated root brings in a logarithm (x^r and x^r ln x); a complex pair a +/- bi yields x^a cos(b ln x) and x^a sin(b ln x). The constants are then chosen to fit conditions.
The trial y = x^r is the Cauchy-Euler counterpart of the characteristic-root method, and learning to see WHEN it applies — only when the coefficient powers line up with derivative orders — is half the skill. The honest caveat about repeated and complex roots: they do NOT just give more powers. A double root forces a logarithm, x^r ln x, exactly as a double exponential root forced an x factor; and complex roots produce oscillation in ln x, not in x. Watching for these special cases keeps you from writing down too few independent solutions.
For x^2 y'' + 3x y' + y = 0, trying y = x^r gives r(r-1) + 3r + 1 = r^2 + 2r + 1 = (r + 1)^2 = 0, a double root r = -1. So the two solutions are x^(-1) and x^(-1) ln x, and the general solution is y = (c1 + c2 ln x) / x.
y = x^r reduces a Cauchy-Euler equation to the indicial equation for r.
A repeated indicial root gives x^r and x^r ln x, not x^r and x^(r+1): the second solution is a logarithm, not another power — this is the most common slip.