Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/172

que l’on peut encore changer en ceux-ci

${\begin{aligned}&4nf_{1}\mu _{1}\chi _{2}{\frac {\mu _{2}}{\mathrm {M} _{2}^{2}}}{\widehat {\Gamma }}_{1}(a_{1},a_{2})\mathrm {x} _{2}\cos(\mu _{2}-\mu _{1})t\\+&4nf_{1}\mu _{1}\chi _{3}{\frac {\mu _{3}}{\mathrm {M} _{3}^{2}}}{\widehat {\Gamma }}_{1}(a_{1},a_{3})\mathrm {x} _{3}\cos(\mu _{3}-\mu _{1})t\\+&4nf_{1}\mu _{1}\chi _{4}{\frac {\mu _{4}}{\mathrm {M} _{4}^{2}}}{\widehat {\Gamma }}_{1}(a_{1},a_{4})\mathrm {x} _{4}\cos(\mu _{4}-\mu _{1})t\\&4nf_{1}\mu _{1}\chi _{2}{\frac {\mu _{2}(\mu _{2}-\mu _{1})}{\mathrm {M} _{2}^{2}}}{\widehat {\Gamma }}_{1}(a_{1},a_{2})\int \mathrm {x} _{2}\sin(\mu _{2}-\mu _{1})tdt\\+&4nf_{1}\mu _{1}\chi _{3}{\frac {\mu _{3}(\mu _{3}-\mu _{1})}{\mathrm {M} _{3}^{2}}}{\widehat {\Gamma }}_{1}(a_{1},a_{3})\int \mathrm {x} _{3}\sin(\mu _{3}-\mu _{1})tdt\\+&4nf_{1}\mu _{1}\chi _{4}{\frac {\mu _{4}(\mu _{4}-\mu _{1})}{\mathrm {M} _{4}^{2}}}{\widehat {\Gamma }}_{1}(a_{1},a_{4})\int \mathrm {x} _{4}\sin(\mu _{4}-\mu _{1})tdt.\end{aligned}}$

Par ce moyen, l’équation $(\mathrm {L} )$ ne contiendra plus que des termes de la forme de

$\mathrm {x} _{2}\cos(\mu _{2}-\mu _{1})t,\quad {\frac {d\mathrm {x} _{2}}{dt}}\sin(\mu _{2}-\mu _{1})t,\quad \int \mathrm {x} _{2}\sin(\mu _{2}-\mu _{1})tdt.$

Je reprends maintenant l’équation

${\frac {d^{2}\mathrm {x} _{1}}{dt^{2}}}+\mathrm {M} _{1}^{2}\mathrm {x} _{1}=0,$

laquelle, étant rapportée au second satellite, devient

${\frac {d^{2}\mathrm {x} _{2}}{dt^{2}}}+\mathrm {M} _{2}^{2}\mathrm {x} _{2}=0\,;$

je multiplie cette dernière par $\sin(\mu _{2}-\mu _{1})tdt\,;$ je l’intègre, j’ai

$\int {\frac {d^{2}\mathrm {x} _{2}}{dt^{2}}}\sin(\mu _{2}-\mu _{1})tdt+\mathrm {M} _{2}^{2}\int \mathrm {x} _{2}\sin(\mu _{2}-\mu _{1})tdt=0\,;$

je change l’expression

$\int {\frac {d^{2}\mathrm {x} _{2}}{dt^{2}}}\sin(\mu _{2}-\mu _{1})tdt$