Page:Annales de mathématiques pures et appliquées, 1822-1823, Tome 13.djvu/90

86

INTÉGRALES

 Les corrections sont expliquées en page de discussion

$0=\left\{{\begin{aligned}&M\left[\left({\frac {\operatorname {d} K}{\operatorname {d} y}}\right)_{1}\left({\frac {\operatorname {d} L}{\operatorname {d} z}}\right)_{1}-\left({\frac {\operatorname {d} K}{\operatorname {d} z}}\right)_{1}\left({\frac {\operatorname {d} L}{\operatorname {d} y}}\right)_{1}-\right]\\\\+&N\left[\left({\frac {\operatorname {d} K}{\operatorname {d} z}}\right)_{1}\left({\frac {\operatorname {d} L}{\operatorname {d} x}}\right)_{1}-\left({\frac {\operatorname {d} K}{\operatorname {d} x}}\right)_{1}\left({\frac {\operatorname {d} L}{\operatorname {d} z}}\right)_{1}-\right]\\\\+&\left[\left({\frac {\operatorname {d} K}{\operatorname {d} x}}\right)_{1}\left({\frac {\operatorname {d} L}{\operatorname {d} y}}\right)_{1}-\left({\frac {\operatorname {d} K}{\operatorname {d} y}}\right)_{1}\left({\frac {\operatorname {d} L}{\operatorname {d} z}}\right)_{1}-\right]\end{aligned}}\right\}.$

mais, en différentiant le système des deux équations $K_{1}=0,\ L_{1}=0,$ on a

${\begin{aligned}&\left({\frac {\operatorname {d} K}{\operatorname {d} x}}\right)_{1}{\frac {\operatorname {d} x_{1}}{\operatorname {d} z_{1}}}+\left({\frac {\operatorname {d} K}{\operatorname {d} y}}\right)_{1}{\frac {\operatorname {d} y_{1}}{\operatorname {d} z_{1}}}+\left({\frac {\operatorname {d} K}{\operatorname {d} z}}\right)_{1}=0,\\\\&\left({\frac {\operatorname {d} L}{\operatorname {d} x}}\right)_{1}{\frac {\operatorname {d} x_{1}}{\operatorname {d} z_{1}}}+\left({\frac {\operatorname {d} L}{\operatorname {d} y}}\right)_{1}{\frac {\operatorname {d} y_{1}}{\operatorname {d} z_{1}}}+\left({\frac {\operatorname {d} L}{\operatorname {d} z}}\right)_{1}=0\,;\end{aligned}}$

d’où, on conclut

${\begin{aligned}&\left[\left({\frac {\operatorname {d} K}{\operatorname {d} x}}\right)_{1}\left({\frac {\operatorname {d} L}{\operatorname {d} y}}\right)_{1}-\left({\frac {\operatorname {d} K}{\operatorname {d} y}}\right)_{1}\left({\frac {\operatorname {d} L}{\operatorname {d} x}}\right)_{1}\right]{\frac {\operatorname {d} x_{1}}{\operatorname {d} z_{1}}}\\\\=&\left[\left({\frac {\operatorname {d} K}{\operatorname {d} y}}\right)_{1}\left({\frac {\operatorname {d} L}{\operatorname {d} z}}\right)_{1}-\left({\frac {\operatorname {d} K}{\operatorname {d} z}}\right)_{1}\left({\frac {\operatorname {d} L}{\operatorname {d} y}}\right)_{1}\right],\\\\&\left[\left({\frac {\operatorname {d} K}{\operatorname {d} x}}\right)_{1}\left({\frac {\operatorname {d} L}{\operatorname {d} y}}\right)_{1}-\left({\frac {\operatorname {d} K}{\operatorname {d} y}}\right)_{1}\left({\frac {\operatorname {d} L}{\operatorname {d} x}}\right)_{1}\right]{\frac {\operatorname {d} y_{1}}{\operatorname {d} z_{1}}}\\\\=&\left[\left({\frac {\operatorname {d} K}{\operatorname {d} z}}\right)_{1}\left({\frac {\operatorname {d} L}{\operatorname {d} x}}\right)_{1}-\left({\frac {\operatorname {d} K}{\operatorname {d} x}}\right)_{1}\left({\frac {\operatorname {d} L}{\operatorname {d} z}}\right)_{1}\right],\end{aligned}}$

au moyen de quoi l’équation ci-dessus, en $M$ et $N,$ deviendra par substitution et suppression du facteur commun,