= Solution
The <fundamental theorem of calculus> first gives $y'(x)=1+\int_0^x f(y(s))\,ds$. Integrating again, or differentiating the proposed expression twice and checking its initial values, yields the <Volterra integral equation>
$$
\boxed{y(x)=x+\int_0^x(x-s)f(y(s))\,ds.}
$$
For $f(y)=-y$, the <initial value problem> is solved by $\boxed{y(x)=\sin x}$. Its <Picard iteration> constructs the sine series directly. The base approximation is $y_0(x)=x$. If
$$
y_n(x)=\sum_{k=0}^n(-1)^k\frac{x^{2k+1}}{(2k+1)!},
$$
then elementary integration gives
$$
\int_0^x(x-s)\frac{s^{2k+1}}{(2k+1)!}\,ds=\frac{x^{2k+3}}{(2k+3)!}.
$$
Substituting this finite sum in the next update, then shifting its index, proves the same formula with $n+1$ in place of $n$. This is the required <mathematical induction>. The resulting partial sums converge to $\boxed{\sin x}$ for every real $x$, and uniformly on every bounded interval: the absolute terms on $|x|\le R$ are bounded by the convergent series $\sum_kR^{2k+1}/(2k+1)!$.
The printed update lists $n=1,2,\ldots$, which would leave $y_1$ unspecified if read literally. The usual <Picard iteration>, and the requested induction from $y_0$, use the update starting at $n=0$; that intended convention is used here.
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