the Cauchy-Euler equation
/ koh-SHEE OY-ler /
Almost all the slick methods so far needed constant coefficients. But one special VARIABLE-coefficient equation is just as solvable, and it appears all over physics and engineering — wherever a problem has a natural scaling symmetry, like fields around a point or stresses in a wedge. That equation is the Cauchy-Euler equation, and its hallmark is that the power of x in each coefficient matches the order of the derivative it multiplies.
In standard form the (second-order) Cauchy-Euler equation is a x^2 y'' + b x y' + c y = 0, with constants a, b, c. Notice the pattern: x^2 sits with y'', x^1 with y', x^0 with y. The nth-order version continues the same way, a_n x^n y^(n) + ... + a_1 x y' + a_0 y = 0. This precise matching is what makes the equation 'equidimensional' — it is invariant under rescaling x by a constant — and it is exactly why a power y = x^r is the right trial: differentiating x^r and multiplying by the matching power of x reproduces x^r times a number, so the equation collapses to an algebraic (indicial) equation for r.
Cauchy-Euler equations matter because they are the simplest variable-coefficient equations you can solve in closed form, and they arise naturally from Laplace's equation in polar or spherical coordinates and from problems with scale symmetry. Two honest cautions. First, the point x = 0 is singular — it is a regular singular point — so solutions are typically stated for x > 0 (or you work with |x|). Second, do not over-generalize: replacing x^2 by (x - 1)^2 etc. still works after a shift, but an equation whose coefficient powers do NOT line up with the derivative orders is simply not Cauchy-Euler and the x^r trick will not apply.
Solve x^2 y'' - 2x y' + 2y = 0 for x > 0. Trying y = x^r gives x^r times r(r-1) - 2r + 2 = x^r times (r^2 - 3r + 2) = 0, so r^2 - 3r + 2 = 0 with roots r = 1 and r = 2. The general solution is y = c1 x + c2 x^2.
When x-powers match derivative orders, the trial y = x^r turns the ODE into an algebraic equation for r.
The indicial equation r(r-1) - ... is NOT just r^2 - ...: the y'' term contributes r(r-1) because differentiating x^r twice brings down r then r-1 — forgetting that factor is the classic Cauchy-Euler error.