Page:Annales de mathématiques pures et appliquées, 1814-1815, Tome 5.djvu/277

271

D’ASTRONOMIE.

$-{\frac {2\pi \operatorname {d} t}{q\operatorname {d} \lambda }}={\frac {pu^{2}v(2-xy)}{\left(pu^{3}-qr^{2}\right)}},$

on demande ${\frac {\operatorname {d} ^{2}t}{\operatorname {d} \lambda ^{2}}}\,?$

Faisant

$z={\frac {u^{2}v(2-xy)}{pu^{3}-qr^{2}}},$

on aura

$-{\frac {2\pi \operatorname {d} t}{q\operatorname {d} \lambda }}=pz\,;$

donc

${\frac {\operatorname {d} ^{2}t}{\operatorname {d} \lambda ^{2}}}=-{\frac {pq}{2\pi }}.{\frac {\operatorname {d} z}{\operatorname {d} \lambda }},$

reste donc à trouver ${\frac {\operatorname {d} z}{\operatorname {d} \lambda }}.$

Comparant à la formule générale

${\frac {u^{m}vr^{\varepsilon }s}{\left(pu^{3}-qr^{2}\right)^{n}}},$

on trouve

${\begin{array}{lll}&m=2,&s=2-xy,\\&\varepsilon =0,&M=-y,\\&n=1,&N=-x,\\\end{array}}$

on en tire

${\begin{aligned}F&=r(-5y+x^{2}y+6xy^{2}-2x^{2}y^{3}),\\G&=2r(2x-y),\\H&=r(-2x+y+x^{2}y),\\\end{aligned}}$

donc

${\frac {\operatorname {d} z}{\operatorname {d} \lambda }}={\frac {1}{\left(pu^{3}-qr^{2}\right)^{3}}}\left\{{\begin{aligned}&p^{2}u^{7}v(-5y+x^{2}y+6xy^{2}-2x^{2}y^{3})\\&-2pqu^{4}r^{2}v(2x-y)\\&+q^{2}ur^{4}v\left(-4x+3y+3x^{2}y-xy^{2}-x^{3}y^{2}\right)\\\end{aligned}}\right\}\,;$

donc