= Solution
{c}
The product $(\tanh s)y(s)$ is continuous, so the <fundamental theorem of calculus> makes the integral continuously differentiable. Differentiating the integral equation and then evaluating at zero gives $y'=(\tanh x)y$, $y(0)=1$. Since $(\log\cosh x)'=\tanh x$, an <integrating factor> is $1/\cosh x$. Thus $(y/\cosh x)'=0$, and
$$
\boxed{y(x)=\cosh x.}
$$
Put $L(x)=\log\cosh x$, so $L(0)=0$ and $L'=\tanh x$. The claimed <Picard iterate> formula holds for $n=0$. Assuming it at $n$, termwise integration of the finite sum gives
$$
y_{n+1}(x)=1+\sum_{k=0}^n\frac1{k!}\int_0^xL(s)^kL'(s)ds
=1+\sum_{k=0}^n\frac{L(x)^{k+1}}{(k+1)!}.
$$
This proves by induction
$$
\boxed{y_n(x)=\sum_{k=0}^n\frac{(\log\cosh x)^k}{k!}\longrightarrow e^{\log\cosh x}=\cosh x.}
$$
The same calculation works for negative $x$ because it uses an antiderivative rather than a monotonicity substitution. On every bounded interval $|x|\leq R$, $0\leq L(x)\leq\log\cosh R$, so comparison with the exponential-series tail also proves uniform convergence there.
Back to article page
