Page:Lagrange - Œuvres (1867) vol. 1.djvu/694

place de $\int {\frac {dz^{2}}{dt^{2}}}dy,$ et ${\mathfrak {B}}_{5}\int (\cos \mathrm {H} t)z'dz$ à la place de $\int \Psi dz,$ on aura, après les réductions,

$(o)\left\{{\begin{aligned}{\frac {d^{2}y}{dt^{2}}}&+\left[h^{2}+i(6h\mathrm {f-2g} )\right]y^{2}+2[\mathrm {g} -2h\mathrm {f} +\mathrm {\left(f^{2}-B\right)} +\mathrm {n} {\mathfrak {A}}^{2}]y\\&+\mathrm {A} +i(\mathrm {g} -2h\mathrm {f} )z^{2}-2ih^{2}y^{3}+ih^{2}yz^{2}+2iz{\frac {dydz}{dt^{2}}}+2\mathrm {n} \int \Phi dy=0,\end{aligned}}\right.$

$(p)\quad {\frac {dz^{2}}{dt^{2}}}+\left(h^{2}+2ih\mathrm {f} \right)z^{2}+\mathrm {B} -4i\mathrm {B} y-4ih^{2}yz^{2}+2i\mathrm {n} {\mathfrak {B}}_{5}\int (\cos \mathrm {H} t)z'dz=0.$

Ces substitutions faites, on trouvera, en ordonnant les termes,

${\begin{aligned}{\frac {d^{2}y}{dt^{2}}}=&\left[-h^{2}-i\left(6h\mathrm {f-2g} +3i\mathrm {f} ^{2}+\mathrm {n} {\mathfrak {P}}_{4}\right)\right.\\&-6i\beta (\mathrm {g} -2h\mathrm {f} -i\mathrm {f} ^{2}+{\frac {2}{3}}i\mathrm {B+n} {\mathfrak {A}}_{2})\\&+\left.8i^{2}\gamma \mathrm {B} -6i^{2}\delta \mathrm {A} -2i^{2}\varepsilon \mathrm {B} +2i^{2}\eta h^{2}\mathrm {B} \right]y\\&-\left(\mathrm {g} -2h\mathrm {f} -i\mathrm {f^{2}+n} {\mathfrak {A}}_{2}\right)-2i\beta \mathrm {A} -2i\gamma \mathrm {B} +2i^{2}\eta (\mathrm {g} -2h\mathrm {f} )\mathrm {B} \\&+\left[3ih^{2}+3i^{2}(4h\mathrm {f-g} )-4i\beta \left[h^{2}+i(6h\mathrm {f-2g} )\right]-15i^{2}\delta (\mathrm {g} -2h\mathrm {f} )\right]y^{2}\\&+\left[{\frac {3}{2}}i\left(h^{2}+i\mathrm {g} \right)-2i^{2}\beta (\mathrm {g} -2h\mathrm {f} )-4i\gamma (h^{2}+2ih\mathrm {f} )\right.\\&\qquad \qquad \qquad \qquad -\left.i^{2}\varepsilon (g-2hf)-4i^{2}\varepsilon h^{2}(g-2hf)\right]z^{2}\\&+\left(-6i^{2}h^{2}+10i^{2}\beta h^{2}-9i^{2}\delta h^{2}\right)y^{3}\\&+\left(-3i^{2}h^{2}+i^{2}\beta h^{2}+16i^{2}\gamma h^{2}-5i^{2}\varepsilon h^{2}+4i^{2}\eta h^{4}\right)yz^{2}\\&+\left(-4i^{2}\beta -4i^{2}\gamma +4i^{2}\varepsilon -5i^{2}\eta h^{2}\right)z{\frac {dydz}{dt^{2}}}\\&+n[-\Phi -2i\beta \Phi y-4i\beta \int \Phi dy-4i^{2}\gamma {\mathfrak {B}}_{3}\int (\cos \mathrm {H} t)z'dz],\\\\{\frac {d^{2}z}{dt^{2}}}=&\left[-h^{2}-i\left(2h\mathrm {f} +i\mathrm {f} ^{2}+\mathrm {n} {\mathfrak {A}}_{4}\right)-i\mu (\mathrm {g} -2h\mathrm {f} -i\mathrm {f} ^{2}+\mathrm {n} {\mathfrak {A}}_{2})i\right.\\&+2i\nu \left(\mathrm {g} -2h\mathrm {f} -i\mathrm {f} ^{2}+\mathrm {n} {\mathfrak {A}}_{2}\right)\left(h^{2}+2ih\mathrm {f} \right)\\&\qquad \qquad -\left.3i^{2}\nu h^{2}(2\mathrm {A+B} )-6i^{2}\pi \mathrm {B} -2i^{2}\rho \mathrm {A} +2i^{2}\sigma h^{2}\mathrm {A} \right]z\\&+\left[4i\left(h^{2}+2ih\mathrm {f} \right)-2i\mu \left[h^{2}+i(4h\mathrm {f-g} )\right]\right.\\&+2i\nu \left[h^{2}+i(6h\mathrm {f-2g} )\right]\left(h^{2}+2ih\mathrm {f} \right)-20i^{2}\nu h^{2}(\mathrm {g} -2h\mathrm {f} )\\&\quad \qquad \qquad \qquad \qquad \qquad -\left.6i^{2}\rho (g-2hf)+6i^{2}\sigma h^{2}(g-2hf)\right]zy\end{aligned}}$