Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/180

${\begin{aligned}&-{\frac {n}{2}}f_{2}\chi _{4}{\overset {\circ }{\Psi }}_{1}(a_{2},a_{4})\ \mathrm {x} _{4}\ \left[\cos(\mu _{4}-\mu _{1})t+\cos(\mu _{4}-2\mu _{2}+\mu _{1})t\right]\\&-{\frac {n}{2}}f_{2}\chi _{4}{\overset {\circ }{\Pi }}_{1}(a_{2},a_{4}){\frac {d\mathrm {x} _{4}}{dt}}\left[\sin(\mu _{4}-\mu _{1})t+\sin(\mu _{4}-2\mu _{2}+\mu _{1})t\right]=0,\end{aligned}}$

équation que je réduis, par la raison que j’ai dite tantôt, à celle-ci

${\begin{aligned}{\frac {d^{2}\mathrm {x} _{2}}{dt^{2}}}&\cos(\mu _{2}-\mu _{1})t+\mathrm {M} _{2}^{2}\mathrm {x} _{2}\cos(\mu _{2}-\mu _{1})t-{\frac {n}{2}}f_{2}\chi _{1}{\overset {\circ }{\Psi }}_{1}(a_{2},a_{1})\mathrm {x} _{1}\\-&{\frac {n}{2}}f_{2}\chi _{3}{\overset {\circ }{\Psi }}_{1}(a_{2},a_{3})\mathrm {x} _{3}\cos(\mu _{3}-\mu _{1})t-{\frac {n}{2}}f_{2}\chi _{3}{\overset {\circ }{\Pi }}_{1}(a_{2},a_{3}){\frac {d\mathrm {x} _{3}}{dt}}\sin(\mu _{3}-\mu _{1})t\\-&{\frac {n}{2}}f_{2}\chi _{4}{\overset {\circ }{\Psi }}_{1}(a_{2},a_{4})\mathrm {x} _{4}\cos(\mu _{4}-\mu _{1})t-{\frac {n}{2}}f_{2}\chi _{4}{\overset {\circ }{\Pi }}_{1}(a_{2},a_{4}){\frac {d\mathrm {x} _{4}}{dt}}\sin(\mu _{4}-\mu _{1})t=0,\end{aligned}}$

laquelle me donne, après les substitutions,

(2^o).

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

En troisième lieu, je multiplie l’équation $(\mathrm {M} _{3})$ par $\sin(\mu _{3}-\mu _{1})t\,;$ j’aurai, après les réductions et les substitutions,

(3^o).

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