In $x^{2}y'' + axy' + by = 0$, each derivative comes with a matching power of $x$. Such equations appear in problems with circular or spherical symmetry, and they are the model for the singular points of the next lesson. Instead of $e^{rx}$, as with real roots, the right guess is a power of $x$.
Key idea
For $x^{2}y'' + axy' + by = 0$ on $x > 0$, try $y = x^{r}$: the indicial equation is $$r(r - 1) + ar + b = 0.$$ Distinct real roots: $y = c_1x^{r_1} + c_2x^{r_2}$. A double root $r$: $y = (c_1 + c_2\ln x)x^{r}$. Complex roots $\alpha \pm \beta i$: $y$ $= x^{\alpha}\left(c_1\cos(\beta\ln x) + c_2\sin(\beta\ln x)\right)$.
Why it works
With $y = x^{r}$: $xy' = rx^{r}$ and $x^{2}y'' = r(r - 1)x^{r}$, so every term is a multiple of $x^{r}$ and the equation becomes $\left(r(r - 1) + ar + b\right)x^{r} = 0$. The substitution $x = e^{s}$ explains the other cases: by the chain rule $xy' = \frac{dy}{ds}$ and $x^{2}y'' = \frac{d^{2}y}{ds^{2}} - \frac{dy}{ds}$, so the equation becomes $\frac{d^{2}y}{ds^{2}} + (a - 1)\frac{dy}{ds} + by = 0$, with constant coefficients. Its characteristic equation is exactly the indicial equation. A double root gives $se^{rs} = (\ln x)x^{r}$, and complex roots give $e^{\alpha s}\cos\beta s = x^{\alpha}\cos(\beta\ln x)$.
Worked example: Distinct roots
Solve $x^{2}y'' - 2xy' + 2y = 0$.
- $r(r - 1) - 2r + 2$ $= r^{2} - 3r + 2$ $= (r - 1)(r - 2) = 0$.
Answer: $y = c_1x + c_2x^{2}$
Worked example: A double root, with conditions
Solve $x^{2}y'' + 3xy' + y = 0$, $y(1) = 2$, $y'(1) = -1$.
- $r(r - 1) + 3r + 1$ $= r^{2} + 2r + 1 = (r + 1)^{2}$: $r = -1$ twice, so $y = x^{-1}(c_1 + c_2\ln x)$.
- $y(1) = c_1 = 2$. $y'$ $= -x^{-2}(c_1 + c_2\ln x) + c_2x^{-2}$, so $y'(1) = -c_1 + c_2 = -1$ and $c_2 = 1$.
Answer: $y = \frac{2 + \ln x}{x}$
Worked example: Complex roots
Solve $x^{2}y'' + xy' + 4y = 0$.
- $r(r - 1) + r + 4 = r^{2} + 4 = 0$: $r = \pm 2i$, so $\alpha = 0$, $\beta = 2$.
Answer: $y$ $= c_1\cos(2\ln x) + c_2\sin(2\ln x)$, which oscillates faster and faster as $x \to 0^{+}$
Common mistake
Using $r^{2} + ar + b = 0$ as the indicial equation, as if it were a constant-coefficient equation. The second derivative contributes $r(r - 1)$, not $r^{2}$: for $x^{2}y'' + 3xy' + y = 0$ the roots are $-1, -1$, not those of $r^{2} + 3r + 1$.