Page:Joseph Louis de Lagrange - Œuvres, Tome 12.djvu/393

385

FRAGMENTS.

Faisant encore $\mathrm {A=B} ,$ on aura

${\begin{alignedat}{3}&\mathrm {A} {\frac {d^{2}\mathrm {L} }{dt^{2}}}&+&\mathrm {{\frac {3}{r^{5}}}(C-A)(\zeta '''y-\eta '''z)\nu } &=&0,\\&\mathrm {A} {\frac {d^{2}\mathrm {M} }{dt^{2}}}&+&\mathrm {{\frac {3}{r^{5}}}(C-A)(\xi '''z-\zeta '''x)\nu } &=&0,\\&\mathrm {A} {\frac {d^{2}\mathrm {N} }{dt^{2}}}&+&\mathrm {{\frac {3}{r^{5}}}(C-A)(\eta '''x-\xi '''y)\nu } &=&0,\end{alignedat}}$

 

fin du tome douzième.