Page:Annales de mathématiques pures et appliquées, 1810-1811, Tome 1.djvu/261

253

RÉSOLUES.

Si, dans ces équations, on considère $p,q,r,s,t,$ comme inconnues, on tirera d’abord des deux premières

$p={\frac {{\frac {\mathrm {d} y}{\mathrm {d} v}}{\frac {\mathrm {d} z}{\mathrm {d} u}}-{\frac {\mathrm {d} y}{\mathrm {d} u}}{\frac {\mathrm {d} z}{\mathrm {d} v}}}{{\frac {\mathrm {d} x}{\mathrm {d} u}}{\frac {\mathrm {d} y}{\mathrm {d} v}}-{\frac {\mathrm {d} x}{\mathrm {d} v}}{\frac {\mathrm {d} y}{\mathrm {d} u}}}},\quad q={\frac {{\frac {\mathrm {d} x}{\mathrm {d} u}}{\frac {\mathrm {d} z}{\mathrm {d} v}}-{\frac {\mathrm {d} x}{\mathrm {d} v}}{\frac {\mathrm {d} z}{\mathrm {d} u}}}{{\frac {\mathrm {d} x}{\mathrm {d} u}}{\frac {\mathrm {d} y}{\mathrm {d} v}}-{\frac {\mathrm {d} x}{\mathrm {d} v}}{\frac {\mathrm {d} y}{\mathrm {d} u}}}}$ ;

posant alors, pour abréger,

${\tfrac {\mathrm {d} y}{\mathrm {d} v}}{\tfrac {\mathrm {d} z}{\mathrm {d} u}}-{\tfrac {\mathrm {d} y}{\mathrm {d} u}}{\tfrac {\mathrm {d} z}{\mathrm {d} v}}=\mathrm {G} ,\quad {\tfrac {\mathrm {d} x}{\mathrm {d} u}}{\tfrac {\mathrm {d} z}{\mathrm {d} v}}-{\tfrac {\mathrm {d} x}{\mathrm {d} v}}{\tfrac {\mathrm {d} z}{\mathrm {d} u}}=\mathrm {H} ,\quad {\tfrac {\mathrm {d} x}{\mathrm {d} u}}{\tfrac {\mathrm {d} y}{\mathrm {d} v}}-{\tfrac {\mathrm {d} x}{\mathrm {d} v}}{\tfrac {\mathrm {d} y}{\mathrm {d} u}}=\mathrm {K} .$

auquel cas les valeurs de $p$ et $q$ deviennent

$p=\mathrm {\frac {G}{K}} ,\quad q=\mathrm {\frac {H}{K}} ,$

on tirera des trois dernières équations

$r={\frac {1}{\mathrm {K} ^{3}}}\left\{{\begin{array}{ll}-\mathrm {H} \left\{\left({\frac {\mathrm {d} y}{\mathrm {d} v}}\right)^{2}{\frac {\mathrm {d} ^{2}y}{\mathrm {d} u^{2}}}-2{\frac {\mathrm {d} y}{\mathrm {d} v}}{\frac {\mathrm {d} y}{\mathrm {d} u}}{\frac {\mathrm {d} ^{2}y}{\mathrm {d} u\mathrm {d} v}}+\left({\frac {\mathrm {d} y}{\mathrm {d} u}}\right)^{2}{\frac {\mathrm {d} ^{2}y}{\mathrm {d} v^{2}}}\right\}\\+\mathrm {K} \left\{\left({\frac {\mathrm {d} y}{\mathrm {d} v}}\right)^{2}{\frac {\mathrm {d} ^{2}z}{\mathrm {d} u^{2}}}-2{\frac {\mathrm {d} y}{\mathrm {d} v}}{\frac {\mathrm {d} y}{\mathrm {d} u}}{\frac {\mathrm {d} ^{2}z}{\mathrm {d} u\mathrm {d} v}}+\left({\frac {\mathrm {d} y}{\mathrm {d} u}}\right)^{2}{\frac {\mathrm {d} ^{2}z}{\mathrm {d} v^{2}}}\right\}\\-\mathrm {G} \left\{\left({\frac {\mathrm {d} y}{\mathrm {d} v}}\right)^{2}{\frac {\mathrm {d} ^{2}x}{\mathrm {d} u^{2}}}-2{\frac {\mathrm {d} y}{\mathrm {d} v}}{\frac {\mathrm {d} y}{\mathrm {d} u}}{\frac {\mathrm {d} ^{2}x}{\mathrm {d} u\mathrm {d} v}}+\left({\frac {\mathrm {d} y}{\mathrm {d} u}}\right)^{2}{\frac {\mathrm {d} ^{2}x}{\mathrm {d} v^{2}}}\right\}\\\end{array}}\right\},$ $s={\frac {1}{\mathrm {K} ^{3}}}\left\{{\begin{array}{ll}-\mathrm {H} \left\{{\frac {\mathrm {d} x}{\mathrm {d} v}}{\frac {\mathrm {d} y}{\mathrm {d} v}}{\frac {\mathrm {d} ^{2}y}{\mathrm {d} u^{2}}}-\left[{\frac {\mathrm {d} x}{\mathrm {d} u}}{\frac {\mathrm {d} y}{\mathrm {d} v}}+{\frac {\mathrm {d} x}{\mathrm {d} v}}{\frac {\mathrm {d} y}{\mathrm {d} u}}\right]{\frac {\mathrm {d} ^{2}y}{\mathrm {d} u\mathrm {d} v}}+{\frac {\mathrm {d} x}{\mathrm {d} u}}{\frac {\mathrm {d} y}{\mathrm {d} u}}{\frac {\mathrm {d} ^{2}y}{\mathrm {d} v^{2}}}\right\}\\+\mathrm {K} \left\{{\frac {\mathrm {d} x}{\mathrm {d} v}}{\frac {\mathrm {d} y}{\mathrm {d} v}}{\frac {\mathrm {d} ^{2}z}{\mathrm {d} u^{2}}}-\left[{\frac {\mathrm {d} x}{\mathrm {d} u}}{\frac {\mathrm {d} y}{\mathrm {d} v}}+{\frac {\mathrm {d} x}{\mathrm {d} v}}{\frac {\mathrm {d} y}{\mathrm {d} u}}\right]{\frac {\mathrm {d} ^{2}z}{\mathrm {d} u\mathrm {d} v}}+{\frac {\mathrm {d} x}{\mathrm {d} u}}{\frac {\mathrm {d} y}{\mathrm {d} u}}{\frac {\mathrm {d} ^{2}z}{\mathrm {d} v^{2}}}\right\}\\-\mathrm {G} \left\{{\frac {\mathrm {d} x}{\mathrm {d} v}}{\frac {\mathrm {d} y}{\mathrm {d} v}}{\frac {\mathrm {d} ^{2}x}{\mathrm {d} u^{2}}}-\left[{\frac {\mathrm {d} x}{\mathrm {d} u}}{\frac {\mathrm {d} y}{\mathrm {d} v}}+{\frac {\mathrm {d} x}{\mathrm {d} v}}{\frac {\mathrm {d} y}{\mathrm {d} u}}\right]{\frac {\mathrm {d} ^{2}x}{\mathrm {d} u\mathrm {d} v}}+{\frac {\mathrm {d} x}{\mathrm {d} u}}{\frac {\mathrm {d} y}{\mathrm {d} u}}{\frac {\mathrm {d} ^{2}x}{\mathrm {d} v^{2}}}\right\}\\\end{array}}\right\},$

qui en diffère en ce que, dans la sienne, ce sont u et v qui sont considérés comme des fonctions de $x$ et $y,$ tandis qu’ici, au contraire, ce sont ces dernières variables que nous considérons comme des fonctions des premières. (Voyez le Traité de calcul différentiel et de calcul intégral ; tome II, pages 565 et 566.)