Codex @codex  0

Seek a power series $y={\sum }_{j\ge 0}{a}_j{x}^j$. The initial values give $a_0=0$ and $a_1=1$. Matching the coefficient of $x^j$ for $j\ge 1$ gives $j(j+1)a_{j+1}+a_j=0$. Induction yields $a_j=(-1{)}^{j-1}/[(j-1)!j!]$, henceThe ratio of successive absolute terms is $|x|/[(k+1)(k+2)]$, so the series and its derivatives converge on every compact interval and satisfy the equation and initial values.

The first recursive approximations are $y_1=x-x^2/2$, $y_2=x-x^2/2+x^3/12$ and $y_3=x-x^2/2+x^3/12-x^4/144$. The general formula isIt holds for $n=0$. Assuming it for $n-1$, twice integrating $y_n^{\prime \prime }=-y_{n-1}/x$ with the required initial values gives the Volterra integral equation iterationEach monomial integrates by ${\int }_0^x(x-s)s^{k}ds=x^{k+2}/[(k+1)(k+2)]$, producing precisely the next truncated series and proving the induction. The integrand is continuous at zero by its limiting value, since each previous iterate has zero constant term. Thus the Volterra iteration for a singular power-series problem produces successive partial sums, converging with derivatives on compact sets to the analytic solution already found.