étant une fonction quelconque de
et
on aura ces deux équations-ci :
![{\displaystyle {\begin{aligned}{\frac {dt}{\mathrm {T} }}&={\frac {dx}{\sqrt {\alpha +\beta x+\gamma x^{2}}}},\\{\frac {dt}{\mathrm {T} }}&={\frac {dy}{\sqrt {\alpha +\beta y+\gamma y^{2}}}}\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a180c77059b789b4a8ede3907ac37e1296fea64)
d’où l’on tire, en multipliant en croix et quarrant,
![{\displaystyle {\begin{aligned}{\frac {\mathrm {T} ^{2}dx^{2}}{dt^{2}}}=&\alpha +\beta x+\gamma x^{2},\\{\frac {\mathrm {T} ^{2}dy^{2}}{dt^{2}}}=&\alpha +\beta y+\gamma y^{2},\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20d38f106a8e87e09c4baaf3a7be11838febfe03)
de sorte qu’en différentiant, et regardant
comme constante, on aura
![{\displaystyle {\begin{aligned}{\frac {2\mathrm {T} d\mathrm {T} dx+2\mathrm {T} ^{2}d^{2}x}{dt^{2}}}=&\beta +2\gamma x,\\{\frac {2\mathrm {T} d\mathrm {T} dy+2\mathrm {T} ^{2}d^{2}y}{dt^{2}}}=&\beta +2\gamma y,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de9c05755bfe5928b17d3505445b415ab9ec7664)
équations qui, étant ajoutées ensemble, en supposant
![{\displaystyle x+y=p,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f02c3fbc81cec047c7165dcee3b5b98d45f7adb)
donnent celle-ci :
![{\displaystyle {\frac {\mathrm {T} d\mathrm {T} dp+\mathrm {T} ^{2}d^{2}p}{dt^{2}}}=\beta +\gamma p.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fee8ccb3d22093439b8d51f58db619fee51be09e)
Or, soit
![{\displaystyle x-y=q,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25826142209960c39644cb0fd48ec6824a30ce25)
et supposons
une fonction de
et
en sorte que l’on ait
![{\displaystyle d\mathrm {T} =\mathrm {M} dp+\mathrm {N} dq,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a82692cdbd408a623d7b7ed11267f8201798f1b)
on aura
![{\displaystyle {\frac {d\mathrm {T} dp}{dt^{2}}}={\frac {\mathrm {M} dp^{2}}{dt^{2}}}+{\frac {\mathrm {N} dpdq}{dt^{2}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63c3b1e5ba2cdd0339f4ef01eb96d962b2663d51)
mais
![{\displaystyle {\frac {dpdq}{dt^{2}}}={\frac {dx^{2}-dy^{2}}{dt^{2}}}={\frac {\alpha +\beta x+\gamma x^{2}}{\mathrm {T} ^{2}}}-{\frac {\alpha +\beta y+\gamma y^{2}}{\mathrm {T} ^{2}}}={\frac {\beta q+\gamma pq}{\mathrm {T} ^{2}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50b0698f2c94592f9aa51cfcf96b2c631795bbc6)