Page:Annales de mathématiques pures et appliquées, 1824-1825, Tome 15.djvu/72

${\frac {\frac {\operatorname {d} z'}{\operatorname {d} y'}}{{\frac {\operatorname {d} z'}{\operatorname {d} x'}}x'+{\frac {\operatorname {d} z'}{\operatorname {d} y'}}y'-z'}}={\frac {2y'}{x'^{2}+y'^{2}+z'^{2}}}\,;$

c’est-à-dire, en chassant les dénominateurs et supprimant les accens désormais superflus

${\begin{aligned}&\left(y^{2}+z^{2}-x^{2}\right){\frac {\operatorname {d} z}{\operatorname {d} x}}-2xy{\frac {\operatorname {d} z}{\operatorname {d} y}}=-2xz,\\\\&\left(x^{2}+z^{2}-y^{2}\right){\frac {\operatorname {d} z}{\operatorname {d} y}}-2xy{\frac {\operatorname {d} z}{\operatorname {d} x}}=-2yz\,;\end{aligned}}$

équations d’où on tirera

${\frac {\operatorname {d} z}{\operatorname {d} x}}={\frac {2xz}{x^{2}+y^{2}-z^{2}}},\qquad {\frac {\operatorname {d} z}{\operatorname {d} y}}={\frac {2yz}{x^{2}+y^{2}-z^{2}}}\,;$

puis donc qu’on a

$\operatorname {d} z={\frac {\operatorname {d} z}{\operatorname {d} x}}\operatorname {d} x+{\frac {\operatorname {d} z}{\operatorname {d} y}}\operatorname {d} y,$

l’équation différentielle totale de la surface cherchée sera

$\operatorname {d} z={\frac {2z(x\operatorname {d} x+y\operatorname {d} y)}{x^{2}+y^{2}-z^{2}}},$

c’est-à-dire,

$\left(x^{2}+y^{2}+z^{2}\right)\operatorname {d} z=2z(x\operatorname {d} x+y\operatorname {d} y+z\operatorname {d} z).$

ou encore