En quatrième lieu, je multiplie la même équation par cos ( μ 3 − μ 1 ) t , {\displaystyle \cos(\mu _{3}-\mu _{1})t,} et je trouve
En cinquième lieu, je multiplie l’équation ( M 4 ) {\displaystyle (\mathrm {M} _{4})} par sin ( μ 4 − μ 1 ) t ; {\displaystyle \sin(\mu _{4}-\mu _{1})t\,;} j’ai
En sixième et dernier lieu, je multiplie la même équation par cos ( μ 4 − μ 1 ) t , {\displaystyle \cos(\mu _{4}-\mu _{1})t,} et je trouve
et je trouve
![{\displaystyle {\begin{aligned}{\frac {d^{2}\mathrm {Q} }{dt^{2}}}&+2(\mu _{3}-\mu _{1}){\frac {dq}{dt}}+\left[\mathrm {M} _{3}^{2}-(\mu _{3}-\mu _{1})^{2}\right]\mathrm {Q} -{\frac {n}{2}}f_{3}\chi _{1}{\overset {\circ }{\Psi }}_{1}(a_{3},a_{1})\mathrm {x} _{1}\\&-{\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]\mathrm {P} -{\frac {n}{2}}f_{3}\chi _{2}{\overset {\circ }{\Pi }}_{1}(a_{3},a_{2}){\frac {dp}{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]\mathrm {R} -{\frac {n}{2}}f_{3}\chi _{4}{\overset {\circ }{\Pi }}_{1}(a_{3},a_{4}){\frac {dr}{dt}}=0.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/729667d9b06492d41dc1e6d48aa4faa9c4ff73f4)
par 
j’ai
![{\displaystyle {\begin{aligned}{\frac {d^{2}r}{dt^{2}}}&-2(\mu _{4}-\mu _{1}){\frac {d\mathrm {R} }{dt}}+\left[\mathrm {M} _{4}^{2}-(\mu _{4}-\mu _{1})^{2}\right]r+{\frac {n}{2}}f_{4}\chi _{1}{\overset {\circ }{\Psi }}_{1}(a_{4},a_{1}){\frac {d\mathrm {x} _{1}}{dt}}\\&-{\frac {n}{2}}f_{4}\chi _{2}\left[{\overset {\circ }{\Psi }}_{1}(a_{4},a_{2})-(\mu _{2}-\mu _{1}){\overset {\circ }{\Pi }}_{1}(a_{4},a_{2})\right]p+{\frac {n}{2}}f_{4}\chi _{2}{\overset {\circ }{\Pi }}_{1}(a_{4},a_{2}){\frac {d\mathrm {P} }{dt}}\\&-{\frac {n}{2}}f_{4}\chi _{3}\left[{\overset {\circ }{\Psi }}_{1}(a_{4},a_{3})-(\mu _{3}-\mu _{1}){\overset {\circ }{\Pi }}_{1}(a_{4},a_{3})\right]q+{\frac {n}{2}}f_{4}\chi _{3}{\overset {\circ }{\Pi }}_{1}(a_{4},a_{3}){\frac {d\mathrm {Q} }{dt}}=0.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a3304dbd22fd4f4de031d1d4349004f9b7397f2)
et je trouve
![{\displaystyle {\begin{aligned}{\frac {d^{2}\mathrm {R} }{dt^{2}}}&+2(\mu _{4}-\mu _{1}){\frac {dr}{dt}}+\left[\mathrm {M} _{4}^{2}+(\mu _{4}-\mu _{1})^{2}\right]\mathrm {R} -{\frac {n}{2}}f_{4}\chi _{1}{\overset {\circ }{\Psi }}_{1}(a_{4},a_{1})\mathrm {x} _{1}\\&-{\frac {n}{2}}f_{4}\chi _{2}\left[{\overset {\circ }{\Psi }}_{1}(a_{4},a_{2})-(\mu _{2}-\mu _{1}){\overset {\circ }{\Pi }}_{1}(a_{4},a_{2})\right]\mathrm {P} -{\frac {n}{2}}f_{4}\chi _{2}{\overset {\circ }{\Pi }}_{1}(a_{4},a_{2}){\frac {dp}{dt}}\\&-{\frac {n}{2}}f_{4}\chi _{3}\left[{\overset {\circ }{\Psi }}_{1}(a_{4},a_{3})-(\mu _{3}-\mu _{1}){\overset {\circ }{\Pi }}_{1}(a_{4},a_{3})\right]\mathrm {Q} -{\frac {n}{2}}f_{4}\chi _{3}{\overset {\circ }{\Pi }}_{1}(a_{4},a_{3}){\frac {dq}{dt}}=0.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed0da5f5044229133f0f2bbab51552637870213b)
