où l’on aura
![{\displaystyle {\begin{aligned}&p^{2}+q^{2}={\sqrt {1-n^{2}z^{2}}},\\&{\frac {q}{p}}={\frac {{\sqrt {1+nz}}-{\sqrt {1-nz}}}{\left({\sqrt {1+nz}}+{\sqrt {1-nz}}\right){\sqrt {-1}}}}={\frac {nz}{1+{\sqrt {1-n^{2}z^{2}}}}}{\frac {1}{\sqrt {-1}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a87230a6d5a263e346877fa87bb73d2f27ce52b)
De sorte que si l’on fait, pour abréger,
![{\displaystyle \mathrm {Z} ={\frac {z}{1+{\sqrt {1-n^{2}z^{2}}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0988e586422af875fbda3155640878c9d6c40f0)
et qu’on remarque que
![{\displaystyle {\frac {1}{\mathrm {Z} }}{\frac {d\mathrm {Z} }{dz}}={\frac {1}{z{\sqrt {1-n^{2}z^{2}}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca836e197776d52ca96bca7847a75b4f3399464a)
on aura
![{\displaystyle {\frac {1}{z(1+nz\sin t)}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e41d2e705a391b1b2adcfdb7fd0a5c4a9162abeb)
![{\displaystyle {\frac {d\mathrm {Z} }{dz}}\left({\frac {1}{\mathrm {Z} }}-{\frac {n}{\sqrt {-1}}}e^{t{\sqrt {-1}}}+{\frac {n^{2}\mathrm {Z} }{({\sqrt {-1}})^{2}}}e^{2t{\sqrt {-1}}}-{\frac {n^{3}\mathrm {Z} ^{2}}{({\sqrt {-1}})^{3}}}e^{3t{\sqrt {-1}}}+\ldots \right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c54e9f2b109e9bec3e953562d24037d8d0d56ba)
![{\displaystyle +\left.{\frac {n}{\sqrt {-1}}}e^{-t{\sqrt {-1}}}+{\frac {n^{2}\mathrm {Z} }{({\sqrt {-1}})^{2}}}e^{-2t{\sqrt {-1}}}+{\frac {n^{3}\mathrm {Z} ^{2}}{({\sqrt {-1}})^{3}}}e^{-3t{\sqrt {-1}}}+\ldots \right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d34f828452ab1a62b77e70357c681b2fadbe930)
et il ne s’agira plus que de développer, suivant les puissances de
les quantités ![{\displaystyle {\frac {1}{\mathrm {Z} }}{\frac {d\mathrm {Z} }{dz}},{\frac {d\mathrm {Z} }{dz}},\mathrm {Z} {\frac {d\mathrm {Z} }{dz}},\ldots .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d43f3307740f5dabf1efa25f2d849a631c7ffd8)
Considérons, en général, la quantité
![{\displaystyle \mathrm {Z} ^{r-1}{\frac {d\mathrm {Z} }{dz}}={\frac {1}{r}}{\frac {d\mathrm {Z} ^{r}}{dz}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/acdec66ff1e5c10d3f5707220f1fb7eb15c7f8a7)
et faisant
en sorte que
![{\displaystyle \mathrm {Z} ^{r-1}{\frac {d\mathrm {Z} }{dz}}={\frac {1}{r}}{\frac {dy}{dz}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21cdaebc7b8f097a3576bd976ffbb5251f821d9d)
j’aurai
![{\displaystyle {\frac {dy}{y}}={\frac {rd\mathrm {Z} }{\mathrm {Z} }}={\frac {rdz}{z{\sqrt {1-n^{2}z^{2}}}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20daa153dacf3a6658d57b9c7c356160b3dc68a3)
d’où je tire, en multipliant les deux membres de cette équation en croix et différentiant après les avoir carrés,
![{\displaystyle r^{2}ydz^{2}-\left(1-2n^{2}z^{2}\right)zdydz-\left(1-n^{2}z^{2}\right)z^{2}d^{2}y=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ece861b9b1d7302d9634ce1fd9aeb72dbbe81c03)