Page:Lagrange - Œuvres (1867) vol. 1.djvu/649

la cinquième par ${\mathfrak {N}}e^{\rho t}dt,$ on les ajoutera ensemble, et on en prendra l’intégrale en faisant disparaître de dessous le signe $\int$ les différences des variables $y,u,\mathrm {U} ,\ldots \,;$ après quoi on chassera les expressions intégrales $\int ye^{\rho t}dt,\int ue^{\rho t}dt,\int \mathrm {U} e^{\rho t}dt,\ldots ,$ en égalant à zéro leurs coefficients, ce qui donnera

${\begin{aligned}&\left(\rho ^{2}+\mathrm {K} ^{2}\right)\lambda +i\left({\frac {\mathrm {M} }{2}}\mu -{\frac {\mathrm {N} }{2}}{\mathfrak {M}}\rho \right)=0,\\&i\mathrm {\left(M-NH\right)} \lambda +\mathrm {\left(\rho ^{2}+K^{2}-H^{2}\right)} \mu +2H{\mathfrak {M}}\rho =0,\\-&i\mathrm {N\lambda \rho -2H\mu \rho +\left(\rho ^{2}+K^{2}-H^{2}\right)} {\mathfrak {M}}=0,\\&i\mathrm {\left({\frac {M}{2}}-NH\right)} \mu +{\frac {i\mathrm {N} }{2}}{\mathfrak {M}}\rho +\mathrm {\left(\rho ^{2}+K^{2}-4H^{2}\right)} \nu +4\mathrm {H} {\mathfrak {N}}\rho =0,\\-&{\frac {i\mathrm {N} }{2}}\mu \rho +i\mathrm {\left({\frac {M}{2}}-NH\right)} {\mathfrak {M}}-4\mathrm {H} \nu \rho +\mathrm {\left(\rho ^{2}+K^{2}-4H^{2}\right)} {\mathfrak {N}}=0.\end{aligned}}$

De cette manière on aura l’équation intégrale

${\begin{aligned}&\left[\lambda {\frac {dy}{dt}}+\mu {\frac {du}{dt}}+{\mathfrak {M}}{\frac {d\mathrm {U} }{dt}}+\nu {\frac {dv}{dt}}+{\mathfrak {N}}{\frac {d\mathrm {V} }{dt}}+\left(-\lambda \rho +{\frac {i\mathrm {N} }{2}}{\mathfrak {M}}\right)y\right.\\&\qquad +(-\mu \rho -2\mathrm {H} {\mathfrak {M}})u+\left(i\mathrm {N} \lambda +2\mathrm {H} \mu -{\mathfrak {M}}wr\right)\mathrm {U} \\&\qquad +\left.\left(-{\frac {i\mathrm {N} }{2}}{\mathfrak {M}}-\nu \rho -4\mathrm {H} {\mathfrak {N}}\right)\nu +\left({\frac {i\mathrm {N} }{2}}\mu +4\mathrm {H} \nu -{\mathfrak {N}}\rho \right)\mathrm {V} \right]e^{\rho t}\\&=\int \mathrm {T} \left(\lambda +\mu \cos \mathrm {H} t+{\mathfrak {M}}\sin \mathrm {H} t+\nu \cos 2\mathrm {H} t+{\mathfrak {N}}\sin 2\mathrm {H} t\right)e^{\rho t}dt.\end{aligned}}$

Soit $\rho ^{2}=-\mathrm {R} ^{2},$ de sorte que $\rho =\mathrm {R} {\sqrt {-1}},$ et

$\mu =i\alpha \lambda ,\quad \nu =i^{2}\beta \lambda ,\quad {\mathfrak {M}}=i{\mathfrak {A}}\rho \lambda ,\quad {\mathfrak {N}}=i^{2}{\mathfrak {B}}\rho \lambda ,$

et l’on aura premièrement

${\begin{aligned}-&\mathrm {R^{2}+K^{2}} +i^{2}\mathrm {\left({\frac {M}{2}}\alpha +{\frac {N}{2}}R^{2}\alpha \right)} =0,\\&\mathrm {M\ \,-NH+\left(-R^{2}+K^{2}-H^{2}\right)\alpha -2HR^{2}} {\mathfrak {A}}=0,\\-&\mathrm {N\,\ \ -2H\alpha +\left(-R^{2}+K^{2}-H^{2}\right)} {\mathfrak {A}}=0,\end{aligned}}$