Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 8.djvu/454

$+{\frac {2i^{2}\left(i^{2}-1\right)}{\left(1+i^{2}\right)^{2}}}$

$+\left\{{\begin{aligned}&\alpha e\ \cos(nt+\theta )\\-&\alpha e'\cos(n't+\theta ')\end{aligned}}\right\}\left\{{\begin{aligned}&\left[\left({\frac {db}{dq}}\right)-{\frac {1}{2}}\left({\frac {db_{2}}{dq}}\right)\right]\sin(n't-nt+\mathrm {B} )\\+{\frac {1}{2}}&\left[({\frac {db_{1}}{dq}})-\ \ \left({\frac {db_{3}}{dq}}\right)\right]\sin 2(n't-nt+\mathrm {B} )\\+\ \ &\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \end{aligned}}\right\}$

$+{\frac {15i^{3}}{4\left(1+i^{2}\right)^{2}}}\alpha ^{2}ee'\left[(b)-{\frac {1}{2}}(b_{2})\right]\sin \mathrm {V} +{\frac {3}{4}}\alpha ^{2}\gamma \gamma '{\frac {i^{2}}{1+i^{2}}}\left[(b)-{\frac {1}{2}}(b_{2})\right]\sin \mathrm {U}$

${\begin{aligned}-2\alpha ^{2}ee'{\frac {i^{2}\left(1-3i^{2}+i^{4}\right)}{\left(1+i^{2}\right)^{3}}}&\left[\left({\frac {db}{dq}}\ \right)-{\frac {1}{2}}\left({\frac {db_{2}}{dq}}\ \ \right)\right]\sin \mathrm {V} \\+{\frac {\alpha ^{2}\gamma \gamma 'i^{3}}{\left(1+i^{2}\right)^{2}}}&\left[\left({\frac {db}{dq}}\ \right)-{\frac {1}{2}}\left({\frac {db_{2}}{dq}}\ \ \right)\right]\sin \mathrm {U} \\-{\frac {2\alpha ^{2}ee'i^{3}\left(1-i^{2}\right)^{2}}{\left(1+i^{2}\right)^{4}}}&\left[\left({\frac {d^{2}b}{dq^{2}}}\right)-{\frac {1}{2}}\left({\frac {d^{2}b_{2}}{dq^{2}}}\right)\right]\sin \mathrm {V} .\end{aligned}}$

Soit

${\begin{aligned}{\frac {15i^{3}}{4\left(1+i^{2}\right)^{2}}}\left[(b)-{\frac {1}{2}}(b_{2})\right]-{\frac {2i^{2}\left(1-3i^{2}+i^{4}\right)}{\left(1+i^{2}\right)^{3}}}&\left[\left({\frac {db}{dq}}\ \ \right)-{\frac {1}{2}}\left({\frac {db_{2}}{dq}}\ \right)\right]\\-{\frac {i^{3}\left(1-i^{2}\right)^{2}}{\left(1+i^{2}\right)^{4}}}&\left[\left({\frac {d^{2}b}{dq^{2}}}\right)-{\frac {1}{2}}\left({\frac {d^{2}b_{2}}{dq^{2}}}\right)\right]=\mathrm {K} ,\\{\frac {3i^{2}}{4\left(1+i^{2}\right)}}\left[(b)-{\frac {1}{2}}(b_{2})\right]+{\frac {i^{3}}{\left(1+i^{2}\right)^{2}}}&\left[\left({\frac {db}{dq}}\ \ \right)-{\frac {1}{2}}\left({\frac {db_{2}}{dq}}\ \right)\right]=\mathrm {L} \,;\end{aligned}}$

${\begin{aligned}\mathrm {C} =&\ \ i\left[\quad (b\ )-{\frac {1}{2}}(b_{2})\right],\\\sideset {^{1}}{}{\mathrm {C} }=&2i\left[{\frac {1}{2}}(b_{1})-{\frac {1}{2}}(b_{3})\right],\\\sideset {^{2}}{}{\mathrm {C} }=&3i\left[{\frac {1}{2}}(b_{2})-{\frac {1}{2}}(b_{4})\right],\\\ldots &\ldots \ldots \ldots \ldots \,;\end{aligned}}$

${\begin{aligned}\mathrm {D} =&{\frac {i\left(i^{2}-2\right)}{2\left(1+i^{2}\right)}}\left[\quad (b\ )-{\frac {1}{2}}(b_{2})\right]+{\frac {i^{2}\left(i^{2}-1\right)}{\left(1+i^{2}\right)^{2}}}\left[\quad \left({\frac {db}{dq}}\ \ \right)-{\frac {1}{2}}\left({\frac {db_{2}}{dq}}\right)\right],\\\sideset {^{1}}{}{\mathrm {D} }=&{\frac {i\left(i^{2}-2\right)}{2\left(1+i^{2}\right)}}\left[{\frac {1}{2}}(b_{1})-{\frac {1}{2}}(b_{3})\right]+{\frac {i^{2}\left(i^{2}-1\right)}{\left(1+i^{2}\right)^{2}}}\left[{\frac {1}{2}}\left({\frac {db_{1}}{dq}}\right)-{\frac {1}{2}}\left({\frac {db_{3}}{dq}}\right)\right],\\\sideset {^{2}}{}{\mathrm {D} }=&{\frac {i\left(i^{2}-2\right)}{2\left(1+i^{2}\right)}}\left[{\frac {1}{2}}(b_{2})-{\frac {1}{2}}(b_{4})\right]+{\frac {i^{2}\left(i^{2}-1\right)}{\left(1+i^{2}\right)^{2}}}\left[{\frac {1}{2}}\left({\frac {db_{2}}{dq}}\right)-{\frac {1}{2}}\left({\frac {db_{4}}{dq}}\right)\right],\\\ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \,;\\\mathrm {E} =&{\frac {i^{2}\left(i^{2}-1\right)}{\left(1+i^{2}\right)^{2}}}\left[\quad \left({\frac {db}{dq}}\ \,\right)-{\frac {1}{2}}\left({\frac {db_{2}}{dq}}\right)\right]-{\frac {i\left(1-2i^{2}\right)}{2\left(1+i^{2}\right)}}\left[\quad (b\ )-{\frac {1}{2}}(b_{2})\right],\\\sideset {^{1}}{}{\mathrm {E} }=&{\frac {i^{2}\left(i^{2}-1\right)}{\left(1+i^{2}\right)^{2}}}\left[{\frac {1}{2}}\left({\frac {db_{1}}{dq}}\right)-{\frac {1}{2}}\left({\frac {db_{3}}{dq}}\right)\right]-{\frac {i\left(1-2i^{2}\right)}{2\left(1+i^{2}\right)}}\left[{\frac {1}{2}}(b_{1})-{\frac {1}{2}}(b_{3})\right],\\\ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \,;\\\end{aligned}}$