Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/436

donc, faisant varier $f,a,i$ et $u,$ on en tirera la valeur de $\delta u,$ laquelle sera

$\delta u={\frac {-3t{\sqrt {\cfrac {2(1+m)}{a^{3}}}}{\cfrac {\delta a}{a}}+\sin u\,\delta f+\delta i}{1-f\cos u}}.$

Substituant donc cette valeur de $\delta u$ dans les expressions précédentes de $\delta x,\delta y,\delta z,$ on aura les valeurs cherchées, lesquelles seront évidemment de cette forme

${\begin{aligned}\delta x=&\mathrm {A} \delta a+\mathrm {B} \delta f+\mathrm {C} \delta g+\mathrm {D} \delta i,\\\delta y=&\mathrm {E} \delta a\,+\mathrm {F} \delta f+\mathrm {G} \delta g+\mathrm {H} \delta i,\\\delta z=&\mathrm {K} \delta b\,+\mathrm {L} \delta c,\end{aligned}}$

en supposant, pour abréger,

${\begin{aligned}\mathrm {A} =&{\frac {\cos u-f}{2}}+{\frac {3t}{2}}{\sqrt {\frac {2(1+m)}{a^{3}}}}{\frac {\sin u}{1-f\cos u}},\\\mathrm {B} =&-{\frac {a}{2}}\left(1+{\frac {\sin ^{2}u}{1-f\cos u}}\right),\\\mathrm {C} =&-{\frac {a{\sqrt {1-f^{2}}}}{2f}}\sin u,\\\mathrm {D} =&-{\frac {a}{2}}{\frac {\sin u}{1-f\cos u}},\\\mathrm {E} =&{\frac {\sqrt {1-f^{2}}}{2}}\sin u-{\frac {3t}{2}}{\sqrt {\frac {2(1+m)}{a^{3}}}}{\frac {{\sqrt {1-f^{2}}}\cos u}{1-f\cos u}},\\\mathrm {F} =&-{\frac {af}{2{\sqrt {1-f^{2}}}}}\sin u+{\frac {a{\sqrt {1-f^{2}}}}{2}}{\frac {\sin u\cos u}{1-f\cos u}},\\\mathrm {G} =&{\frac {a}{2f}}(\cos u-f),\\\mathrm {H} =&{\frac {a{\sqrt {1-f^{2}}}}{2}}{\frac {\cos u}{1-f\cos u}},\\\mathrm {K} =&{\frac {a}{2}}(\cos u-f),\\\mathrm {L} =&{\frac {a{\sqrt {1-f^{2}}}}{2}}\sin u.\end{aligned}}$