Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 9.djvu/385

deux équations

${\frac {\partial p}{\partial t}}=\mathrm {A} ,\qquad {\frac {\partial q}{\partial t}}=\mathrm {M}$

et en substituant les valeurs de $p$ et de $q$ que l’on en tirera dans la quantité $\mathrm {X} ,$ ou $p\sin ht+q\cos ht+\mathrm {S} ,$ ce qui revient à la règle que nous avons donnée, article IV. L’équation $(a')$ de l’article III, par exemple, comparée à

$y=\mathrm {X} +\mathrm {Y} t+\ldots ,$

donne

${\begin{aligned}\mathrm {X} =&p\sin t+q\cos t\\&+{\frac {\alpha mp}{16}}\left(1+{\frac {3\alpha m}{16}}\right)\sin 3t+{\frac {\alpha mq}{16}}\left(1-{\frac {3\alpha m}{16}}\right)\cos 3t\\&\qquad \qquad \qquad \qquad \qquad \qquad +{\frac {\alpha ^{2}m^{2}p}{768}}\sin 5t+{\frac {\alpha ^{2}m^{2}q}{768}}\cos 5t,\\\end{aligned}}$

${\begin{aligned}\mathrm {Y} =-{\frac {\alpha mq}{4}}\left(1+{\frac {3\alpha m}{16}}\right)\sin t&-{\frac {\alpha mp}{4}}\left(1-{\frac {3\alpha m}{16}}\right)\cos t\\&-{\frac {\alpha ^{2}m^{2}q}{64}}\sin 3t-{\frac {\alpha ^{2}m^{2}p}{64}}\cos 3t\,;\end{aligned}}$

l’équation

$\mathrm {Y} ={\frac {\partial \mathrm {X} }{\partial p}}{\frac {\partial p}{\partial \theta }}+\ldots$

donnera conséquemment la suivante

${\begin{aligned}-{\frac {\alpha mq}{4}}\left(1+{\frac {3\alpha m}{16}}\right)\sin t&-{\frac {\alpha mp}{4}}\left(1-{\frac {3\alpha m}{16}}\right)\cos t\\&-{\frac {\alpha ^{2}m^{2}q}{64}}\sin 3t-{\frac {\alpha ^{2}m^{2}p}{64}}\cos 3t\end{aligned}}$

${\begin{aligned}=&{\frac {\partial p}{\partial \theta }}\sin t+{\frac {\partial q}{\partial \theta }}\cos t\\&+{\frac {\alpha m}{16}}{\frac {\partial p}{\partial \theta }}\left(1+{\frac {3\alpha m}{16}}\right)\sin 3t+{\frac {\alpha m}{16}}{\frac {\partial q}{\partial \theta }}\left(1-{\frac {3\alpha m}{16}}\right)\cos 3t\\&\qquad \qquad \qquad \qquad \qquad \qquad +{\frac {\alpha ^{2}m^{2}}{768}}{\frac {\partial p}{\partial \theta }}\sin 5t+{\frac {\alpha ^{2}m^{2}}{768}}{\frac {\partial q}{\partial \theta }}\cos 5t,\end{aligned}}$

d’où l’on tire, en comparant les coefficients de $\sin t$ et de $\cos t$ et en y changeant $\theta$ en $t,$

${\begin{aligned}{\frac {dp}{dt}}=&-{\frac {\alpha mq}{4}}\left(1+{\frac {3\alpha m}{16}}\right),\\{\frac {dq}{dt}}=&-{\frac {\alpha mp}{4}}\left(1-{\frac {3\alpha m}{16}}\right),\end{aligned}}$