Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/169

Le texte de cette page a été corrigé et est conforme au fac-similé.

les termes dont il s’agit viennent uniquement des termes qui renferment

{\begin{array}{rrr}x_{2}\cos(\mu _{2}-\mu _{1})t,&x_{3}\cos(\mu _{3}-\mu _{1})t,&x_{4}\cos(\mu _{4}-\mu _{1})t,\\y_{2}\sin(\mu _{2}-\mu _{1})t,&y_{3}\sin(\mu _{3}-\mu _{1})t,&y_{4}\sin(\mu _{4}-\mu _{1})t,\end{array}}

{\begin{array}{rrr}\int x_{2}\sin(\mu _{2}-\mu _{1})tdt,&\int x_{3}\sin(\mu _{3}-\mu _{1})tdt,&\int x_{4}\sin(\mu _{4}-\mu _{1})tdt,\\\int y_{2}\cos(\mu _{2}-\mu _{1})tdt,&\int y_{3}\cos(\mu _{3}-\mu _{1})tdt,&\int y_{4}\cos(\mu _{4}-\mu _{1})tdt,\end{array}}

en tant qu’on y substitue $\mathrm {x_{2},x_{3},x_{4}\,;\ y_{2},y_{3},y_{4}} ,$ à la place de $x_{2},x_{3},x_{4}\,;$ $y_{2},y_{3},y_{4},$ de sorte qu’on pourra réduire cette équation à celle-ci

{\begin{aligned}(\mathrm {L} )\qquad {\frac {d^{2}x_{1}}{dt^{2}}}&+M_{1}^{2}\mathrm {x} _{1}\\&-nf_{1}\chi _{2}{\breve {\Psi }}_{1}(a_{1},a_{2})\mathrm {x} _{2}\cos(\mu _{2}-\mu _{1})t\\&-nf_{1}\chi _{3}{\breve {\Psi }}_{1}(a_{1},a_{3})\mathrm {x} _{3}\cos(\mu _{3}-\mu _{1})t\\&-nf_{1}\chi _{4}{\breve {\Psi }}_{1}(a_{1},a_{4})\mathrm {x} _{4}\cos(\mu _{4}-\mu _{1})t\\&+nf_{1}\chi _{2}{\breve {\Gamma }}_{1}\,(a_{1},a_{2})\mathrm {y} _{2}\sin(\mu _{2}-\mu _{1})t\\&+nf_{1}\chi _{3}{\breve {\Gamma }}_{1}\,(a_{1},a_{3})\mathrm {y} _{3}\sin(\mu _{3}-\mu _{1})t\\&+nf_{1}\chi _{4}{\breve {\Gamma }}_{1}\,(a_{1},a_{4})\mathrm {y} _{4}\sin(\mu _{4}-\mu _{1})t\\&+2nf_{1}\mu _{1}\chi _{2}{\widehat {\Psi }}_{1}(a_{1},a_{2})\int \mathrm {x} _{2}\sin(\mu _{2}-\mu _{1})tdt\\&+2nf_{1}\mu _{1}\chi _{3}{\widehat {\Psi }}_{1}(a_{1},a_{3})\int \mathrm {x} _{3}\sin(\mu _{3}-\mu _{1})tdt\\&+2nf_{1}\mu _{1}\chi _{4}{\widehat {\Psi }}_{1}(a_{1},a_{4})\int \mathrm {x} _{4}\sin(\mu _{4}-\mu _{1})tdt\\&+2nf_{1}\mu _{1}\chi _{2}{\widehat {\Gamma }}_{1}\,(a_{1},a_{2})\int \mathrm {y} _{2}\cos(\mu _{2}-\mu _{1})tdt\\&+2nf_{1}\mu _{1}\chi _{3}{\widehat {\Gamma }}_{1}\,(a_{1},a_{3})\int \mathrm {y} _{3}\cos(\mu _{3}-\mu _{1})tdt\\&+2nf_{1}\mu _{1}\chi _{4}{\widehat {\Gamma }}_{1}\,(a_{1},a_{4})\int \mathrm {y} _{4}\cos(\mu _{4}-\mu _{1})tdt=0.\end{aligned}}