que l’on peut encore changer en ceux-ci
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c69c81c176de76513300d46e771e526968fb6189)
Par ce moyen, l’équation
ne contiendra plus que des termes de la forme de
![{\displaystyle \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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9dc48b6db0b47e52fc63d3a3cb36006a00d78a5c)
Je reprends maintenant l’équation
![{\displaystyle {\frac {d^{2}\mathrm {x} _{1}}{dt^{2}}}+\mathrm {M} _{1}^{2}\mathrm {x} _{1}=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ad3fc1935f3e0d8d29f3129986fbc6159d7937cd)
laquelle, étant rapportée au second satellite, devient
![{\displaystyle {\frac {d^{2}\mathrm {x} _{2}}{dt^{2}}}+\mathrm {M} _{2}^{2}\mathrm {x} _{2}=0\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a8e72f146fa8d391739b0b3a4c6673bec771f93)
je multiplie cette dernière par
je l’intègre, j’ai
![{\displaystyle \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\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae9704938b4d754ae346d28dc0c34a6451416fda)
je change l’expression
![{\displaystyle \int {\frac {d^{2}\mathrm {x} _{2}}{dt^{2}}}\sin(\mu _{2}-\mu _{1})tdt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/272993954fb3635018b4444b4f6ae6dd3b835af4)