Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/370

vante

${\begin{aligned}\Phi =&-{\frac {1}{2\nu ^{2}}}\left(1+{\frac {3\varepsilon ^{2}}{2}}+{\frac {9\varepsilon ^{2}}{4}}\cos 2(\xi +\eta )\right)\\&-{\frac {3\varepsilon }{4\nu ^{2}\lambda }}\left[\left(3+{\frac {9\varepsilon ^{2}}{4}}\right)\cos(\xi +\eta )+{\frac {5\varepsilon ^{2}}{2}}\cos 3(\xi +\eta )\right]x^{2}\\&-{\frac {3}{2}}{\frac {\mathrm {B} +\mathrm {C} \cos 2z}{x^{4}}}-2{\frac {\mathrm {D} \sin z+\mathrm {E} \sin 3z}{x^{5}}},\\\\\Pi =&-{\frac {9\varepsilon ^{2}}{8\nu ^{2}}}\sin 2(\xi +\eta )x\\&-{\frac {3\varepsilon }{4\nu ^{2}\lambda }}\left[\left(1+{\frac {3\varepsilon ^{2}}{4}}\right)\sin(\xi +\eta )+{\frac {5\varepsilon ^{2}}{4}}\sin 3(\xi +\eta )\right]x^{2}\\&-{\frac {\mathrm {C} \sin 2z}{x^{4}}}-{\frac {\mathrm {D} \cos z+3\mathrm {E} \cos 3z}{2x^{5}}}.\end{aligned}}$

26. On substituera maintenant ces valeurs de $\Phi$ et de $\Pi$ dans l’équation VI de l’orbite de la Lune, laquelle deviendra par là, à cause de $x={\frac {1}{u}},$

${\begin{aligned}\mathrm {VIII} .\quad 2{\frac {d^{2}u}{d\varphi ^{2}}}&+u-{\frac {1}{k^{2}}}+{\frac {1}{2\nu ^{2}k^{2}u^{3}}}\left[1+{\frac {3\varepsilon ^{2}}{2}}+{\frac {9\varepsilon ^{2}}{4}}\cos 2(\xi +\eta )\right]\\&+{\frac {3\varepsilon }{4\nu ^{2}k^{2}\lambda u^{4}}}\left[\left(3+{\frac {9\varepsilon ^{2}}{4}}\right)\cos(\xi +\eta )+{\frac {5\varepsilon ^{2}}{2}}\cos 3(\xi +\eta )\right]\\&+{\frac {3u^{2}}{2k^{2}}}(\mathrm {B+C} \cos 2z)+{\frac {2u^{3}}{k^{2}}}(\mathrm {D} \sin z+\mathrm {E} \sin 3z)\\&-{\frac {9\varepsilon ^{2}}{4k^{4}\nu ^{2}}}\int {\frac {\sin 2(\xi +\eta )}{u^{4}}}d\varphi -{\frac {9\varepsilon ^{2}}{8\nu ^{2}}}{\frac {\sin 2(\xi +\eta )d\omega }{u^{4}d\varphi }}\\&-{\frac {3\varepsilon }{2k^{4}\nu ^{2}\lambda }}\left[\left(1+{\frac {3\varepsilon ^{2}}{4}}\right)\int {\frac {\sin(\xi +\eta )}{u^{5}}}d\varphi -{\frac {5\varepsilon ^{2}}{4}}{\frac {\sin 3(\xi +\eta )}{u^{5}}}d\varphi \right]\\&-{\frac {3\varepsilon }{2k^{4}\nu ^{2}\lambda }}\left[\left(1+{\frac {3\varepsilon ^{2}}{4}}\right){\frac {\sin(\xi +\eta )du}{u^{5}d\varphi }}-{\frac {5\varepsilon ^{2}}{4}}{\frac {\sin 3(\xi +\eta )}{u^{5}d\varphi }}\right]\\&+{\frac {2}{k^{4}}}\left(\mathrm {C} \int u\sin 2zd\varphi -{\frac {\mathrm {D} }{2}}\int u^{2}\cos zd\varphi -{\frac {3\mathrm {E} }{2}}\int u^{2}\cos 3zd\varphi \right)\\&+{\frac {1}{k^{2}}}\left(\mathrm {C} {\frac {\sin 2z.udu}{d\varphi }}-{\frac {\mathrm {D} }{2}}{\frac {\cos z.u^{2}du}{d\varphi }}-{\frac {3\mathrm {E} }{2}}{\frac {\cos 3z.u^{2}du}{d\varphi }}\right)=0.\end{aligned}}$