Solution
= Solution
For the <Euler-Cauchy equation>, substitute $y=x^n$ on an interval not containing zero. The coefficient of $x^{n-2}$ is
$$
n(n-1)-3n+3=n^2-4n+3=(n-1)(n-3).
$$
The distinct roots give
$$
\boxed{y(x)=C_1x+C_2x^3.}
$$
These solutions are linearly independent since their <Wronskian> is $2x^3\ne0$ away from zero. Both nevertheless approach zero as $x\to0$, which will make the endpoint conditions in the next part insufficient for uniqueness.