à la place de
on aura les équations suivantes
![{\displaystyle {\begin{aligned}d\,\delta h=&-\mu {\frac {\sqrt {2h}}{1+m}}(x\mathrm {Y} -y\mathrm {X} )dt,\\d\,\delta a=&-{\frac {\mu }{1+m}}a^{2}(\mathrm {X} dx+\mathrm {Y} dy),\\d\,\delta f=&-{\frac {\mu ydt}{\sqrt {2h(1+m)}}}\left[\left(f+{\frac {x}{r}}\right)\mathrm {X} +{\frac {y}{r}}\mathrm {Y} \right]+{\frac {2dyd\,\delta h}{dt{\sqrt {2h(1+m)}}}},\\d\,\delta g=&\quad {\frac {\mu xdt}{\sqrt {2h(1+m)}}}\left[\left(f+{\frac {x}{r}}\right)\mathrm {X} +{\frac {y}{r}}\mathrm {Y} \right]-{\frac {2dxd\,\delta h}{dt{\sqrt {2h(1+m)}}}},\\d\,\delta b=&\quad {\frac {\mu y}{\sqrt {2h(1+m)}}}\mathrm {Z} dt,\\d\,\delta c=&-{\frac {\mu x}{\sqrt {2h(1+m)}}}\mathrm {Z} dt,\\d\,\delta i=&3t{\sqrt {\frac {2(1+m)}{a^{3}}}}{\frac {d\,\delta a}{a}}-{\frac {2r^{2}d\,\delta g}{f{\sqrt {a^{3}h}}}}+{\frac {y(fx-4h)d\,\delta f}{2{\sqrt {ah^{3}}}}}\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1fb4944a734aa200ef273be65e11bc61b074dbc)
et l’équation entre
étant différentiée aussi, donnera
![{\displaystyle ad\,\delta h-hd\,\delta a+{\frac {a^{2}fd\,\delta f}{2}}=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc9be984c7b53e2b5f144b39616febbf05e8d8a9)
qui servira à déterminer, si l’on veut,
en connaissant
et ![{\displaystyle d\,\delta a.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94259102e81ef6c536fc4809f82c877ee2e57042)
Or je remarque qu’on a cette combinaison
![{\displaystyle xd\,\delta f+yd\,\delta g=2{\frac {xdy-ydx}{dt{\sqrt {2h(1+m)}}}}d\,\delta h=2d\,\delta h\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4784f7f3f696b3e53519c73b8ebf6f0408fb8a6)
de sorte qu’on aura
![{\displaystyle d\,\delta g={\frac {2d\,\delta h-xd\,\delta f}{y}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c48732c3d1df78ced7ae469861bd69cce502bcb7)
ainsi, comme
![{\displaystyle d\,\delta f={\frac {2(hd\,\delta a-ad\,\delta h)}{a^{2}f}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9cd7f98e0d0c18e947bbb14952c42da0a8a87d42)
on aura
![{\displaystyle d\,\delta g={\frac {2}{afy}}\left[(af+x)d\,\delta h-xh\right]{\frac {d\,\delta a}{a}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b7013f5631099939a7116c12f950e62e773f2d3)
valeurs que l’on pourra employer à la place des précédentes.