= Solution
{c}
Work first on $x>0$. For $b=0$, the <power-law ansatz> $y=x^r$ gives the <indicial equation> $r(r-1)+r-a^2=r^2-a^2=0$. Therefore the <Cauchy-Euler equation> has <general solution>
$$
\boxed{y(x)=
\begin{cases}
C_1x^a+C_2x^{-a},&a>0,\\
C_1+C_2\log x,&a=0.
\end{cases}}
$$
For $a=0$ the repeated <indicial root> supplies the <logarithm> rather than two copies of the constant solution. Alternatively, the <change of variables> $t=\log x$ turns the <differential equation> into $Y''-a^2Y=0$. On $x<0$ the corresponding real formulas use $|x|$ and $\log|x|$, with independent constants on the two sides of zero.
For general $b$, normalize the <second-order linear differential equation> as
$$
y''+P(x)y'+Q(x)y=0,\qquad P(x)=\frac1x,\qquad Q(x)=-\frac{a^2}{x^2}+bx^{\beta-2}.
$$
The coefficient $P$ is singular at zero, but $xP(x)=1$ and $x^2Q(x)=-a^2+bx^\beta$ are <analytic functions> there because $\beta$ is a positive integer. The <regular singular point criterion for a second-order equation> therefore proves that zero is a <regular singular point> for every allowed $a,b,\beta$. The <indicial equation> remains $r^2-a^2=0$.
Back to article page
