ピカールの逐次近似法
( t 2 2 ) l {\displaystyle {\begin{aligned}\varphi _{1}(t)&=1+\int _{0}^{t}s\,ds=1+{\frac {t^{2}}{2}}\\\varphi _{2}(t)&=1+\int _{0}^{t}s\left(1+{\frac {s^{2}}{2}}\right)ds=1+{\frac {t^{2}}{2}}+{\frac {t^{4}}{2\cdot 4}}\\&\cdots \\\varphi _{n}(t)&=1+{\frac {t^{2}}{2}}+{\frac {t^{4}}{2\cdot 4}}+\cdots +{\frac {t^{2n}}{2\cdot 4\cdots \cdot (2n)}}=\sum _{l=0}^{n}{\frac {1}{l!}}\left({\frac {t^{2}}{2}}\right)^{l}\end{aligned}}}
よって、解は
φ ( t ) = lim n → ∞ φ n ( t ) = e t 2 / 2 {\displaystyle \varphi (t)=\lim _{n\to \infty }{\varphi _{n}(t)}=e^{t^{2}/2}}
となる。