= Solution
{c}
Near $x=0$, <separation of variables> gives $dy/y=dx/(1-x)$, so $\log|y|=-\log(1-x)+C$. The <initial condition> selects
$$
\boxed{y(x)=\frac1{1-x},\qquad -\infty<x<1.}
$$
Direct substitution verifies both the <ordinary differential equation> and the <initial condition>. The endpoint $x=1$ is excluded because the coefficient and this solution are singular there; the displayed interval is the maximal connected real interval containing the initial point.
For $|x|\leq h<1$ and any two admissible values $y,\widetilde y$,
$$
|f(x,y)-f(x,\widetilde y)|
=\frac{|y-\widetilde y|}{|1-x|}
\leq\frac1{1-h}|y-\widetilde y|.
$$
This is a <Lipschitz condition> in the dependent variable, uniform in $x$, with <Lipschitz constant> $L=(1-h)^{-1}$. Also $|f(x,y)|\leq(1+k)/(1-h)$ on the prescribed rectangle, so the usual local hypotheses of the <Picard-Lindelöf theorem> hold.
Integrating from the initial point gives the <integral equation>
$$
y(x)=1+\int_0^x\frac{y(t)}{1-t}\,dt.
$$
Conversely, differentiating this equation using the <fundamental theorem of calculus> recovers the <initial value problem>. Start <Picard iteration> with the constant initial approximation $y_0(x)=1$ and set
$$
y_{n+1}(x)=1+\int_0^x\frac{y_n(t)}{1-t}\,dt.
$$
The first three approximations are
$$
\boxed{y_0=1,\qquad y_1=1-\log(1-x),\qquad
y_2=1-\log(1-x)+\frac12\log^2(1-x).}
$$
To prove the general expression by <mathematical induction>, put $\ell(x)=\log(1-x)$. The initial approximation gives the formula for $n=0$. If it holds for $n$, then the <substitution rule>
$$
\int_0^x\frac{\ell(t)^k}{1-t}\,dt=-\frac{\ell(x)^{k+1}}{k+1}
$$
gives
$$
y_{n+1}(x)=1+\sum_{k=0}^n\frac{(-1)^{k+1}\ell(x)^{k+1}}{(k+1)!}
=\sum_{k=0}^{n+1}\frac{(-1)^k\ell(x)^k}{k!}.
$$
Thus the formula follows at every stage.
For every fixed real $x<1$, these are partial sums of the <exponential series> evaluated at the finite real number $-\log(1-x)$. Therefore
$$
\boxed{\lim_{n\to\infty}y_n(x)=\exp(-\log(1-x))=\frac1{1-x}
\quad\text{for every }x<1.}
$$
This range is larger than $|x|<1$: the expansion is in powers of $\log(1-x)$, not a <Taylor series> in $x$. For example it converges at $x=-2$. It also converges uniformly on every <compact interval> contained in $(-\infty,1)$, because $|\log(1-x)|$ is bounded there and the <exponential series> supplies a uniform summable bound. At $x=1$ the approximations are undefined, while for $x>1$ the real integral from zero crosses the singularity. This is <Picard iteration for a simple-pole linear differential equation>.
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