De sorte que les valeurs de d Q , d Q ′ {\displaystyle d\mathrm {Q} ,d\mathrm {Q} '} et d Q ″ {\displaystyle d\mathrm {Q} ''} deviendront
Donc, faisant toutes ces substitutions dans les équations ( K ) , {\displaystyle (\mathrm {K} ),} elles deviendront
et 
deviendront
![{\displaystyle {\begin{aligned}d\mathrm {Q} \ \,=&{\frac {2dr}{r^{2}}}+i^{3}\left[-{\frac {d\left(r^{2}\right)}{\mathrm {R} ^{3}}}+{\frac {3z\left[d(\mathrm {R} z)-\sigma dt\right]}{\mathrm {R} ^{4}}}\right]\\&+i^{4}\left[-{\frac {3zd\left(r^{2}\right)}{\mathrm {R} ^{4}}}+{\frac {\left(15z^{2}-3r^{2}\right)[d(\mathrm {R} z)-\sigma dt]}{2\mathrm {R} ^{5}}}\right]\\&+i^{5}\left[-{\frac {\left(15z^{2}-3r^{2}\right)d\left(r^{2}\right)}{2\mathrm {R} ^{5}}}+{\frac {\left(35z^{3}-15zr^{2}\right)[d(\mathrm {R} z)-\sigma dt]}{2\mathrm {R} ^{6}}}\right]+\ldots ,\\\\d\mathrm {Q} '\,=&-{\frac {1}{i}}\left[{\frac {d(\mathrm {R} z)+\sigma dt}{r^{3}}}\right]+i^{2}\left[{\frac {-6zd\mathrm {R} +d(\mathrm {R} z)+\sigma dt}{\mathrm {R} ^{3}}}\right]+\ldots ,\\d\mathrm {Q} ''=&{\frac {1}{i}}\left[{\frac {d(\mathrm {R} z)+\sigma dt}{r^{3}}}\right]-{\frac {2dr}{r^{2}}}-i^{2}\left[{\frac {-6zd\mathrm {R} +d(\mathrm {R} z)+\sigma dt}{\mathrm {R} ^{3}}}\right]+\ldots .\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03b375a59f3263f655f9d98b83034a57670ec1ec)
elles deviendront
![{\displaystyle (g)\left\{{\begin{aligned}&{\frac {d^{2}\left(r^{2}\right)}{2dt^{2}}}\\&-{\frac {\mathrm {A+B} }{r}}-\mathrm {D} \left[{\frac {3z^{2}-r^{2}}{\mathrm {R} ^{3}}}+\int \left(-{\frac {d\left(r^{2}\right)}{\mathrm {R} ^{3}}}+{\frac {3z[d(\mathrm {R} z)-\sigma dt]}{\mathrm {R} ^{4}}}\right)\right]\\&-i\ \mathrm {D} \left[{\frac {15z^{3}-9r^{2}z}{2\mathrm {R} ^{4}}}+\int \left(-{\frac {3zd\left(r^{2}\right)}{\mathrm {R} ^{4}}}+{\frac {\left(15z^{2}-3r^{2}\right)[d(\mathrm {R} z)-\sigma dt]}{2\mathrm {R} ^{5}}}\right)\right]\\&-i^{2}\mathrm {D} \left[{\frac {35z^{4}-30z^{2}r^{2}+3r^{4}}{2\mathrm {R} ^{5}}}\right.\\&\qquad \left.+\int \left(-{\frac {\left(15z^{2}-3r^{2}\right)d\left(r^{2}\right)}{2\mathrm {R} ^{5}}}+{\frac {\left(35z^{3}-15zr^{2}\right)\left[d(\mathrm {R} z)-\sigma dt\right]}{2\mathrm {R} ^{6}}}\right)\right]-\ldots =\mathrm {const} .,\end{aligned}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf91bbf1f63734ff530d6fabca0c98265a1a7ab8)
![{\displaystyle {\begin{aligned}{\frac {d^{2}\left(\mathrm {R} ^{2}\right)}{2i^{2}dt^{2}}}&-{\frac {\mathrm {D} }{i^{2}\mathrm {R} }}+{\frac {\mathrm {B} }{i}}\left[{\frac {\mathrm {R} z}{r^{3}}}+\int {\frac {d(\mathrm {R} z)+\sigma dt}{r^{3}}}\right]-{\frac {i(\mathrm {A+B} )}{\mathrm {R} }}-\ldots =\mathrm {const} .,\\\\{\frac {d^{2}\left(\mathrm {R} ^{2}\right)}{2i^{2}dt^{2}}}&-{\frac {d^{2}(\mathrm {R} z)}{idt^{2}}}+{\frac {d^{2}\left(r^{2}\right)}{2dt^{2}}}-{\frac {\mathrm {D} }{i^{2}\mathrm {R} }}\\&-{\frac {1}{i}}\left[{\frac {\mathrm {D} z}{\mathrm {R} ^{2}}}+\mathrm {A} {\frac {\mathrm {R} z}{r^{3}}}+\mathrm {A} \int {\frac {d(\mathrm {R} z)+\sigma dt}{r^{3}}}\right]\\&-{\frac {\mathrm {D} \left(3z^{2}-r^{2}\right)}{2\mathrm {R} ^{3}}}-{\frac {\mathrm {A} }{r}}-i\left[{\frac {\mathrm {D} \left(5z^{3}-3zr^{2}\right)}{2\mathrm {R} ^{4}}}+{\frac {\mathrm {A+B} }{\mathrm {R} }}\right]-\ldots =\mathrm {const} .,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69a048cc7c2f5808b426b69a862567697d02b360)
