= Solution
{c}
Rewrite the linear equation as $(x^2y)'=0$. The initial value gives the unique solution
$$
\boxed{y(x)=x^{-2}\quad(x>0).}
$$
The <Picard iteration> is its integral equation
$$
y_{n+1}(x)=1-2\int_1^x\frac{y_n(t)}t\,dt.
$$
The proposed formula holds for $n=0$. If it holds at $n$, termwise integration of the finite sum and $d(\log t)/dt=1/t$ give
$$
y_{n+1}(x)
=1+\sum_{k=0}^n\frac{(-2\log x)^{k+1}}{(k+1)!}
=\sum_{k=0}^{n+1}\frac{(-2\log x)^k}{k!}.
$$
This proves the expression by induction. The <exponential series> then gives $y_n(x)\to e^{-2\log x}=x^{-2}$. Convergence is also uniform on every compact subinterval of $(0,\infty)$, since $|\log x|$ is bounded there and the exponential majorant series converges. This is <Picard iteration for a simple-pole linear differential equation>.
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