One Solvable Island in a Sea of Variable Coefficients
Everything in this rung so far has leaned on one luxury: constant coefficients. The whole characteristic-polynomial machine — guess y = e^(rx), read off a degree-n polynomial, find its roots — works only when the numbers in front of y, y', y'', y^(n) do not move. The moment a coefficient depends on x, the lucky guess e^(rx) stops reproducing itself cleanly and the whole method collapses. That is the bad news, and it is the rule: a generic variable-coefficient linear equation has no elementary closed-form solution at all, which is exactly why later rungs reach for power series and numerical methods.
But there is one gorgeous exception — a variable-coefficient equation that is *just as solvable* as a constant-coefficient one. It is the [[ode-cauchy-euler-equation|Cauchy-Euler equation]] (also called the equidimensional or Euler equation). In second order it reads a x^2 y'' + b x y' + c y = 0, with constants a, b, c. Look at the pattern carefully: x^2 sits with y'', x^1 with y', and x^0 with y. The power of x in each coefficient *matches the order of the derivative it multiplies*. That single coincidence is the whole secret.
Why does this exact matching matter so much? Because it makes the equation equidimensional — invariant when you rescale x by a constant. Such scaling symmetry is not a math curiosity: it appears wherever a physical problem has no built-in length scale, like the electric field around a point charge, the temperature around a corner, or the stresses in a wedge. Solving Laplace's equation in polar or spherical coordinates spits out Cauchy-Euler equations directly. So this is not a contrived special case — it is the closed-form heart of a whole family of real problems.
The Right Lucky Guess: y = x^r
For a constant-coefficient equation the lucky guess was y = e^(rx), because the exponential reproduces itself under differentiation. For the Cauchy-Euler equation the right guess is different — a plain power, the [[trial-solution-x-to-the-r|trial solution y = x^r]]. The reason is a perfect bit of matchmaking: differentiating x^r drops the power of x by one each time, and the equation's built-in x-factors push it right back up. Watch the cancellation. If y = x^r then y' = r x^(r-1) and y'' = r(r-1) x^(r-2). Multiply: x^2 y'' becomes r(r-1) x^r, x y' becomes r x^r, and y is just x^r. Every term is now a number times the *same* x^r.
Factor out the common x^r and it cancels, leaving a pure algebraic equation in r with no x anywhere: a r(r-1) + b r + c = 0. This is the [[ode-indicial-equation|indicial equation]] — the Cauchy-Euler counterpart of the characteristic polynomial. It is the same idea as the previous guides, just with a power instead of an exponential. Solve this (degree-n in general) equation for r, and each root hands you a building-block solution x^r. The most common slip lives right here: the y'' term contributes r(r-1), not r^2, because differentiating x^r twice brings down r and then r-1. Forgetting that factor is the classic Cauchy-Euler error.
a x^2 y'' + b x y' + c y = 0 (Cauchy-Euler form) y = x^r -> y' = r x^(r-1) , y'' = r(r-1) x^(r-2) x^2 y'' -> r(r-1) x^r x y' -> r x^r y -> x^r ---------------------------------------------------- indicial equation: a r(r-1) + b r + c = 0
Three Cases for the Roots — With a Logarithmic Twist
The indicial equation has three cases, exactly mirroring the constant-coefficient story you already know — but with one surprise. Distinct real roots r1, r2 are the easy case: two solutions x^(r1) and x^(r2), and the general solution is y = C1 x^(r1) + C2 x^(r2). A repeated root r is where the surprise lives. You might guess the second solution is another power, but it is not — it is x^r ln x. So the general solution is y = (C1 + C2 ln x) x^r. The logarithm here plays exactly the role the extra t did in t e^(rt) for repeated exponential roots; do not write x^r and x^(r+1), the classic over-counting mistake.
Complex conjugate roots alpha ± i beta are the third case, and they too oscillate — but in ln x, not in x. The two real solutions are x^alpha cos(beta ln x) and x^alpha sin(beta ln x), giving y = x^alpha (C1 cos(beta ln x) + C2 sin(beta ln x)). The real part alpha controls how the amplitude grows or shrinks as a power of x; the imaginary part beta sets how fast the solution wiggles against the *logarithmic* clock ln x. So as x runs from 1 to e to e^2, the wave completes equal numbers of swings — it oscillates in geometric, not arithmetic, steps. It is the same Euler's-formula trick from the second-order rung, just read on a logarithmic axis.
The Deeper Reason: Straighten the Axis with x = e^t
Why should logarithms appear, and why does x^r work at all? There is a single change of variable that explains everything: the [[substitution-x-equals-e-to-the-t|substitution x = e^t]], equivalently t = ln x. It stretches the x-axis logarithmically and, in doing so, irons the awkward x-powers perfectly flat. Under the chain rule, x times d/dx becomes d/dt, and x^2 times d^2/dx^2 becomes d^2/dt^2 - d/dt (mind that cross term — dropping the - d/dt is a common slip). Substitute these into a x^2 y'' + b x y' + c y = 0 and *every x cancels*, leaving a plain constant-coefficient equation in t: a y_tt + (b - a) y_t + c y = 0.
Now you are back on home ground. Solve the t-equation with the ordinary characteristic equation, find y as a function of t, then substitute back t = ln x. And watch the magic land: a solution e^(rt) becomes e^(r ln x) = x^r — that *is* the x^r trial, explained. A repeated root's t e^(rt) becomes (ln x) x^r — there are your logarithms. And e^(alpha t) cos(beta t) becomes x^alpha cos(beta ln x) — there is your oscillation in ln x. So the three cases are not three rules to memorize; they are the *same* three constant-coefficient cases, viewed back through a logarithmic window.
Two Worked Examples, Side by Side
Let us run the machine twice — once with the direct x^r trial, once via the substitution — so you trust they agree. Take x^2 y'' - 2x y' + 2y = 0 on x > 0. Here a = 1, b = -2, c = 2.
- Form the indicial equation: r(r-1) - 2r + 2 = r^2 - 3r + 2 = 0. Notice the y'' term gave r(r-1), not r^2.
- Factor: (r - 1)(r - 2) = 0, so r = 1 and r = 2 — two distinct real roots.
- Write the general solution: y = C1 x + C2 x^2. Done — no logarithms because the roots are distinct.
- Cross-check via x = e^t: the equation becomes y_tt - 3 y_t + 2 y = 0, whose characteristic equation r^2 - 3r + 2 = 0 has the same roots. Its y = C1 e^t + C2 e^(2t), and e^t = x, e^(2t) = x^2 — the identical answer.