mais on a
![{\displaystyle {\overset {n}{y}}_{x}+\mathrm {A} {\overset {n}{y}}_{x-1}+\ldots =\mathrm {B} {\overset {1}{y}}_{x}+^{1}\!\mathrm {B} {\overset {1}{y}}_{x-1}+\ldots +\mathrm {X} _{x}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31f0b3a975ec5542c3f2e7ccebb762f41f4e83cc)
donc
![{\displaystyle {\begin{array}{rll}{\overset {1}{y}}_{x}+b_{n}{\overset {1}{y}}_{x-1}+\ldots =&a_{n}\left(\mathrm {B} {\overset {1}{y}}_{x}+^{1}\!\mathrm {B} {\overset {1}{y}}_{x-1}+\ldots \right)&+a_{n}\mathrm {X} _{x}\\&+^{1}\!a_{n}\left(\mathrm {B} {\overset {1}{y}}_{x-1}+\ldots \right)&+^{1}\!a_{n}\mathrm {X} _{x-1}\\&+\ldots \ldots \ldots \ldots &+\ldots \ldots \ldots \\&&+{\overset {n}{u}}_{x},\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48dbb05bb5086cdabe78ee05641082fae203fc33)
et, en ordonnant les différents termes de cette équation,
![{\displaystyle {\overset {1}{y}}_{x}\left(1-a_{n}\mathrm {B} \right)+{\overset {1}{y}}_{x-1}\left(b_{n}-a_{n}\,^{1}\!\mathrm {B} -^{1}\!a_{n}\mathrm {B} \right)+{\overset {1}{y}}_{x-2}\left(^{1}\!b_{n}-a_{n}\,^{2}\!\mathrm {B} -^{1}\!a_{n}\,^{1}\!\mathrm {B} -^{2}\!a_{n}\mathrm {B} \right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/14df64006c4ba641faea7859e11f9dd1d7163b9c)
![{\displaystyle +\ldots -{\overset {n}{u}}_{x}-a_{n}\mathrm {X} _{x}-^{1}\!a_{n}\mathrm {X} _{x-1}-\ldots =0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/950ea9979dcfae075b8b6cf51c31242b42d2255b)
on aura une équation entièrement semblable pour ![{\displaystyle {\overset {2}{y}}_{x},{\overset {3}{y}}_{x},\ldots .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69fbc34181a9c300f730e34468c0080812dfd054)
XVII.
Problème IV. – Je suppose maintenant que les équations rentrantes renferment trois variables, et que l’on ait
![{\displaystyle {\begin{aligned}{\overset {1}{y}}_{x}+\mathrm {A} {\overset {1}{y}}_{x-1}+^{1}\!\mathrm {A} {\overset {1}{y}}_{x-2}+\ldots =&\mathrm {B} {\overset {2}{y}}_{x}+^{1}\!\mathrm {B} {\overset {2}{y}}_{x-1}+^{2}\!\mathrm {B} {\overset {2}{y}}_{x-2}+\ldots \\&+\mathrm {C} {\overset {3}{y}}_{x}+^{1}\!\mathrm {C} {\overset {3}{y}}_{x-1}+^{2}\!\mathrm {C} {\overset {3}{y}}_{x-2}+\ldots \\&+\mathrm {X} _{x},\\\ldots \ldots \ldots \ldots \ldots \ldots &\ldots \ldots \ldots \ldots \ldots \ldots ,\\{\overset {n}{y}}_{x}+\mathrm {A} {\overset {n}{y}}_{x-1}+^{1}\!\mathrm {A} {\overset {n}{y}}_{x-2}+\ldots =&\mathrm {B} {\overset {1}{y}}_{x}+^{1}\!\mathrm {B} {\overset {1}{y}}_{x-1}+^{2}\!\mathrm {B} {\overset {1}{y}}_{x-2}+\ldots \\&+\mathrm {C} {\overset {2}{y}}_{x}+^{1}\!\mathrm {C} {\overset {2}{y}}_{x-1}+\ldots \\&+\mathrm {X} _{x},\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6dd1637a666c4cee9ccada58c215dc9721efb324)
il faut déterminer ![{\displaystyle {\overset {1}{y}}_{x},{\overset {2}{y}}_{x},\ldots .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/092bcf176ffb6589f68d6b5eba39f88d37c80d63)
En suivant le procédé du problème précédent, on arrivera à une