 Solution

The fundamental theorem of calculus first gives $y^{\prime }(x)=1+{\int }_0^{x}f(y(s))ds$. Integrating again, or differentiating the proposed expression twice and checking its initial values, yields the Volterra integral equationFor $f(y)=-y$, the initial value problem is solved by $\boxed{y(x)=\sin x}$. Its Picard iteration constructs the sine series directly. The base approximation is $y_0(x)=x$. Ifthen elementary integration givesSubstituting 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 ${\sum }_k{R}^{2k+1}/(2k+1)!$.

The printed update lists $n=1,2,\dots$, 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.