Page:Annales de mathématiques pures et appliquées, 1822-1823, Tome 13.djvu/81

77

INDÉTERMINÉES.

$\left({\frac {\operatorname {d} S}{\operatorname {d} x}}\right)X+\left({\frac {\operatorname {d} S}{\operatorname {d} y}}\right)Y+\left({\frac {\operatorname {d} S}{\operatorname {d} z}}\right)Z=0.\qquad$(16)

Ajoutant à l’équation (14) le produit de celle-ci par $\lambda ,$ il viendrait

$0=\left\{{\begin{aligned}&\left[\lambda \left({\frac {\operatorname {d} S}{\operatorname {d} x}}\right)+\left({\frac {\operatorname {d} V}{\operatorname {d} x}}\right)-\left({\frac {\operatorname {d} V}{\operatorname {d} x'}}\right)'+\left({\frac {\operatorname {d} V}{\operatorname {d} x''}}\right)''-\left({\frac {\operatorname {d} V}{\operatorname {d} x'''}}\right)'''+\ldots \right]X\\\\+&\left[\lambda \left({\frac {\operatorname {d} S}{\operatorname {d} y}}\right)+\left({\frac {\operatorname {d} V}{\operatorname {d} y}}\right)-\left({\frac {\operatorname {d} V}{\operatorname {d} y'}}\right)'+\left({\frac {\operatorname {d} V}{\operatorname {d} y''}}\right)''-\left({\frac {\operatorname {d} V}{\operatorname {d} y'''}}\right)'''+\ldots \right]Y\\\\+&\left[\lambda \left({\frac {\operatorname {d} S}{\operatorname {d} z}}\right)+\left({\frac {\operatorname {d} V}{\operatorname {d} z}}\right)-\left({\frac {\operatorname {d} V}{\operatorname {d} z'}}\right)'+\left({\frac {\operatorname {d} V}{\operatorname {d} z''}}\right)''-\left({\frac {\operatorname {d} V}{\operatorname {d} z'''}}\right)'''+\ldots \right]Z\end{aligned}}\right\}(17)$

égalant alors séparément à zéro les multiplicateurs de $X,Y,Z$ et éliminant $\lambda$ entre les trois équations résultantes, on obtiendra double équation

$\left.{\begin{aligned}&{\frac {\left({\frac {\operatorname {d} V}{\operatorname {d} x}}\right)-\left({\frac {\operatorname {d} V}{\operatorname {d} x'}}\right)'+\left({\frac {\operatorname {d} V}{\operatorname {d} x''}}\right)''-\left({\frac {\operatorname {d} V}{\operatorname {d} x'''}}\right)'''+\ldots }{\left({\frac {\operatorname {d} S}{\operatorname {d} x}}\right)}}\\\\=&{\frac {\left({\frac {\operatorname {d} V}{\operatorname {d} y}}\right)-\left({\frac {\operatorname {d} V}{\operatorname {d} y'}}\right)'+\left({\frac {\operatorname {d} V}{\operatorname {d} y''}}\right)''-\left({\frac {\operatorname {d} V}{\operatorname {d} y'''}}\right)'''+\ldots }{\left({\frac {\operatorname {d} S}{\operatorname {d} y}}\right)}}\\\\=&{\frac {\left({\frac {\operatorname {d} V}{\operatorname {d} z}}\right)-\left({\frac {\operatorname {d} V}{\operatorname {d} z'}}\right)'+\left({\frac {\operatorname {d} V}{\operatorname {d} z''}}\right)''-\left({\frac {\operatorname {d} V}{\operatorname {d} z'''}}\right)'''+\ldots }{\left({\frac {\operatorname {d} S}{\operatorname {d} z}}\right)}}\end{aligned}}\right\}\,;\ (18)$

laquelle, dans l’hypothèse actuelle, ne devra compter que pour une seule, dont il faudra combiner l’intégrale avec $S=0,$ pour obtenir les valeurs cherchées de $x$ et $y$ en fonction de $z$.