Page:Lagrange - Œuvres (1867) vol. 1.djvu/638

Maintenant on a :

1^o ${\frac {d^{2}u}{dt^{2}}}=2y{\frac {d^{2}y}{dt^{2}}}+2{\frac {dy^{2}}{dt^{2}}}\,;$

donc $2y\times \mathrm {(L)+2(N)}$ donnera, en négligeant les termes de l’ordre de $i,$

${\begin{aligned}{\frac {d^{2}u}{dt^{2}}}&+2\mathrm {A} +6\mathrm {F} y+4\mathrm {G} y^{2}+2\mathrm {H} yz+4\mathrm {H} \int zdy\\&+i\left({\frac {10\mathrm {K} }{3}}y^{3}+2\mathrm {L} y^{2}z+2\mathrm {M} yz^{2}+4\mathrm {L} \int yzdy+4\mathrm {M} \int z^{2}dy\right)=0,\end{aligned}}$

et, en faisant les substitutions précédentes,

(3)

$\left\{{\begin{aligned}{\frac {d^{2}u}{dt^{2}}}&+2\mathrm {A} +6\mathrm {F} y+4\mathrm {G} u+2\mathrm {H} u_{1}+4\mathrm {H} u_{3}\\&+i\left({\frac {10\mathrm {K} }{3}}v+2\mathrm {L} v_{1}+2\mathrm {M} v_{2}+4\mathrm {L} v_{4}+4\mathrm {M} v_{5}\right)=0,\end{aligned}}\right.$

2^o ${\frac {d^{2}u_{1}}{dt^{2}}}=z{\frac {d^{2}y}{dt^{2}}}+y{\frac {d^{2}z}{dt^{2}}}+2{\frac {dydz}{dt^{2}}}\,;$

donc $z\times (\mathrm {L} )+y\times (\mathrm {M} )+2(\mathrm {P} )$ donnera

(4)

$\left\{{\begin{aligned}{\frac {d^{2}u_{1}}{dt^{2}}}&+2\mathrm {C} +3fy+3\mathrm {F} z+2gu+(3\mathrm {G} +h)u_{1}+2\mathrm {H} u_{2}+2(h-\mathrm {G} )u_{3}\\&+i\left[{\frac {5k}{3}}v+(3\mathrm {K} +l)v_{1}+(2\mathrm {L} +m)v_{2}+{\frac {5\mathrm {M} }{3}}v_{3}\right.\\&\qquad \qquad \qquad \qquad \qquad \left.+(2l-4K)v_{4}+(2m-L)v_{5}{\begin{aligned}\\\\\end{aligned}}\right]=0.\end{aligned}}\right.$

3^o ${\frac {d^{2}u_{2}}{dt^{2}}}=2z{\frac {d^{2}z}{dt^{2}}}+2{\frac {dz^{2}}{dt^{2}}}\,;$

donc $2z\times (\mathrm {M} )+2(\mathrm {O} )$ donnera

(5)

$\left\{{\begin{aligned}{\frac {d^{2}u}{dt^{2}}}&+2\mathrm {B} +6fz+6gu_{1}+4hu_{2}-4gu_{3}\\&+i\left(6kv_{1}+4kv_{2}+{\frac {10m}{3}}v_{3}-8kv_{4}-2kv_{5}\right)=0.\end{aligned}}\right.$

4^o ${\frac {d^{2}u_{3}}{dt^{2}}}=z{\frac {d^{2}y}{dt^{2}}}+{\frac {dydz}{dt^{2}}}\,;$