Page:Annales de mathématiques pures et appliquées, 1823-1824, Tome 14.djvu/159

sur quoi il faudra se rappeler de faire varier $z$ dans $P$ et $Q,$ comme fonctions de $x$ et $y.$

Si présentement nous remarquons que

${\begin{aligned}1+Q^{2}+pP&=1+P^{2}+Q^{2}-P(P-p),\\\\1+P^{2}+qQ&=1+P^{2}+Q^{2}-Q(Q-q),\end{aligned}}$

qu’en outre

${\frac {\operatorname {d} q}{\operatorname {d} x}}={\frac {\operatorname {d} p}{\operatorname {d} y}}=s,$

et que, par suite,

${\frac {\operatorname {d} Q}{\operatorname {d} x}}-{\frac {\operatorname {d} P}{\operatorname {d} y}}={\frac {\operatorname {d} (Q-q)}{\operatorname {d} x}}-{\frac {\operatorname {d} (P-p)}{\operatorname {d} y}},$

l’équation $(\delta )$ pourra être écrite ainsi

$(Q-q)\left(P{\frac {\operatorname {d} P}{\operatorname {d} x}}+Q{\frac {\operatorname {d} Q}{\operatorname {d} x}}\right)-(P-p)\left(P{\frac {\operatorname {d} P}{\operatorname {d} y}}+Q{\frac {\operatorname {d} Q}{\operatorname {d} y}}\right)$

$+\left(1+P^{2}+Q^{2}\right)\left\{{\frac {\operatorname {d} (P-p)}{\operatorname {d} y}}-{\frac {\operatorname {d} (Q-q)}{\operatorname {d} x}}\right\}=0\,;$

mais on a

$P{\frac {\operatorname {d} P}{\operatorname {d} x}}+Q{\frac {\operatorname {d} Q}{\operatorname {d} x}}={\frac {1}{2}}.{\frac {\operatorname {d} \left(1+P^{2}+Q^{2}\right)}{\operatorname {d} x}},\quad P{\frac {\operatorname {d} P}{\operatorname {d} y}}+Q{\frac {\operatorname {d} Q}{\operatorname {d} y}}={\frac {1}{2}}.{\frac {\operatorname {d} \left(1+P^{2}+Q^{2}\right)}{\operatorname {d} y}}\,;$

donc notre équation revient encore à

$2\left(1+P^{2}+Q^{2}\right){\frac {\operatorname {d} (Q-q)}{\operatorname {d} x}}-(Q-q){\frac {\operatorname {d} \left(1+P^{2}+Q^{2}\right)}{\operatorname {d} x}}$

$=2\left(1+P^{2}+Q^{2}\right){\frac {\operatorname {d} (P-p)}{\operatorname {d} y}}-(P-p){\frac {\operatorname {d} \left(1+P^{2}+Q^{2}\right)}{\operatorname {d} y}}\,;$