Page:Annales de mathématiques pures et appliquées, 1823-1824, Tome 14.djvu/158

${\begin{array}{ll}1+pP+(z-z'){\frac {\operatorname {d} P}{\operatorname {d} x}}=M,&qP+(z-z'){\frac {\operatorname {d} P}{\operatorname {d} y}}=M',\\\\1+qQ+(z-z'){\frac {\operatorname {d} Q}{\operatorname {d} y}}=N,&pQ+(z-z'){\frac {\operatorname {d} Q}{\operatorname {d} x}}=N',\end{array}}$

on en tirera ainsi

${\begin{array}{ll}M{\frac {\operatorname {d} x}{\operatorname {d} x'}}+M'{\frac {\operatorname {d} y}{\operatorname {d} x'}}=1+P^{2},&M{\frac {\operatorname {d} x}{\operatorname {d} y'}}+M'{\frac {\operatorname {d} y}{\operatorname {d} y'}}=PQ,\\\\N{\frac {\operatorname {d} y}{\operatorname {d} y'}}+N'{\frac {\operatorname {d} x}{\operatorname {d} y'}}=1+Q^{2},&N{\frac {\operatorname {d} y}{\operatorname {d} x'}}+N'{\frac {\operatorname {d} x}{\operatorname {d} x'}}=PQ,\end{array}}$

équations qui donnent

${\begin{array}{ll}{\frac {\operatorname {d} x}{\operatorname {d} x'}}={\frac {\left(1+P^{2}\right)N-PQM'}{MN-M'N'}},&{\frac {\operatorname {d} y}{\operatorname {d} y'}}={\frac {\left(1+Q^{2}\right)M-PQN'}{MN-M'N'}},\\\\{\frac {\operatorname {d} y}{\operatorname {d} x'}}={\frac {PQM-\left(1+P^{2}\right)N'}{MN-M'N'}},&{\frac {\operatorname {d} x}{\operatorname {d} y'}}={\frac {PQN-\left(1+Q^{2}\right)M'}{MN-M'N'}}\,;\end{array}}$

substituant toutes ces valeurs dans l’équation de condition $(\gamma ),$ elle deviendra

$(1+P^{2})\left(N{\frac {\operatorname {d} Q}{\operatorname {d} x}}-N'{\frac {\operatorname {d} Q}{\operatorname {d} y}}\right)-(1+Q^{2})\left(M{\frac {\operatorname {d} Q}{\operatorname {d} y}}-M'{\frac {\operatorname {d} P}{\operatorname {d} x}}\right)$

$=PQ\left(N{\frac {\operatorname {d} P}{\operatorname {d} x}}-N'{\frac {\operatorname {d} P}{\operatorname {d} y}}-M{\frac {\operatorname {d} Q}{\operatorname {d} y}}+M'{\frac {\operatorname {d} Q}{\operatorname {d} x}}\right)\,;$

mettant enfin dans cette dernière, pour $M,N,M',N'$ les fonctions dont ces lettres sont les symboles, on trouvera, toutes réductions faites,

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

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