Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/177

Je substitue ces valeurs dans l’équation $(M_{1}),$ ce qui la change en celle-ci

${\begin{aligned}&{\frac {d^{2}\mathrm {x} _{1}}{dt^{2}}}+\mathrm {M} _{1}^{2}\mathrm {x} _{1}\\&-nf_{1}\chi _{2}\left[{\overset {\circ }{\Psi }}_{1}(a_{1},a_{2})-(\mu _{2}-\mu _{1}){\overset {\circ }{\Pi }}_{1}(a_{1},a_{2})\right]\mathrm {P} -nf_{1}\chi _{2}{\overset {\circ }{\Pi }}_{1}(a_{1},a_{2}){\frac {dp}{dt}}\\&-nf_{1}\chi _{3}\left[{\overset {\circ }{\Psi }}_{1}(a_{1},a_{3})-(\mu _{3}-\mu _{1}){\overset {\circ }{\Pi }}_{1}(a_{1},a_{3})\right]\mathrm {Q} -nf_{1}\chi _{3}{\overset {\circ }{\Pi }}_{1}(a_{1},a_{3}){\frac {dq}{dt}}\\&-nf_{1}\chi _{4}\left[{\overset {\circ }{\Psi }}_{1}(a_{1},a_{4})-(\mu _{4}-\mu _{1}){\overset {\circ }{\Pi }}_{1}(a_{1},a_{4})\right]\mathrm {R} -nf_{1}\chi _{4}{\overset {\circ }{\Pi }}_{1}(a_{1},a_{4}){\frac {dr}{dt}}=0.\end{aligned}}$

C’est l’équation qu’il s’agit maintenant d’intégrer, en regardant les quantités $\mathrm {P,Q,R} ,p,q,r,$ chacune comme une variable particulière. Pour y parvenir, voici comment je m’y prends.

XCIII.

Je reprends les formules

${\begin{aligned}{\frac {d\mathrm {x} _{2}}{dt}}\sin(\mu _{2}-\mu _{1})t=&{\frac {dp}{dt}}-(\mu _{2}-\mu _{1})\mathrm {P} ,\\{\frac {d\mathrm {x} _{3}}{dt}}\sin(\mu _{3}-\mu _{1})t=&{\frac {dq}{dt}}-(\mu _{3}-\mu _{1})\mathrm {Q} ,\\{\frac {d\mathrm {x} _{4}}{dt}}\sin(\mu _{4}-\mu _{1})t=&{\frac {dr}{dt}}-(\mu _{4}-\mu _{1})\mathrm {R} .\end{aligned}}$

et je trouve de même

${\begin{aligned}{\frac {d\mathrm {x} _{2}}{dt}}\cos(\mu _{2}-\mu _{1})t=&{\frac {d\mathrm {P} }{dt}}+(\mu _{2}-\mu _{1})p,\\{\frac {d\mathrm {x} _{3}}{dt}}\cos(\mu _{3}-\mu _{1})t=&{\frac {d\mathrm {Q} }{dt}}+(\mu _{3}-\mu _{1})q,\\{\frac {d\mathrm {x} _{4}}{dt}}\cos(\mu _{4}-\mu _{1})t=&{\frac {d\mathrm {R} }{dt}}+(\mu _{4}-\mu _{1})r.\end{aligned}}$