= Solution
Seek a <power series> $y=\sum_{j\ge0}a_jx^j$. The initial values give $a_0=0$ and $a_1=1$. Matching the coefficient of $x^j$ for $j\ge1$ gives $j(j+1)a_{j+1}+a_j=0$. Induction yields $a_j=(-1)^{j-1}/[(j-1)!j!]$, hence
$$
\boxed{y(x)=\sum_{k=0}^\infty\frac{(-1)^kx^{k+1}}{k!(k+1)!}
=x-\frac{x^2}2+\frac{x^3}{12}-\frac{x^4}{144}+\cdots.}
$$
The 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 is
$$
\boxed{y_n(x)=\sum_{k=0}^n\frac{(-1)^kx^{k+1}}{k!(k+1)!}.}
$$
It holds for $n=0$. Assuming it for $n-1$, twice integrating $y_n''=-y_{n-1}/x$ with the required initial values gives the <Volterra integral equation> iteration
$$
y_n=x-\int_0^x(x-s)\frac{y_{n-1}(s)}s\,ds.
$$
Each 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.
Back to article page
