En comparant ces formules aux formules générales, on aura
![{\displaystyle \mathrm {V} ={\frac {\sqrt {1+y'^{2}}}{\sqrt {z}}}=f(x,y,y',z)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae27de22214c40c8a381fa214c6d3b81f42bd7dd)
et la fonction
qui, étant égalée à zéro, donne l’équation de condition, sera
![{\displaystyle z'-2g+2\varphi (z){\sqrt {1+y'^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef565710c1bd2a1180983529d1ac847f663fec92)
On aura donc
![{\displaystyle \mathrm {\overset {.}{V}} ={\frac {y'{\overset {.}{y}}\,\!'}{{\sqrt {1+y'^{2}}}{\sqrt {z}}}}-{\frac {{\overset {.}{z}}{\sqrt {1+y'^{2}}}}{2z^{\frac {3}{2}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b56e6a87fd3a061c84dfb3f9f4fd0dfb3eeeba1c)
Et de là
![{\displaystyle f'(y')={\frac {y'}{{\sqrt {1+y'^{2}}}{\sqrt {z}}}},\quad f'(z)=-{\frac {\sqrt {1+y'^{2}}}{2z^{\frac {3}{2}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/953cac72aa6e3f098ffabb4049b184688e29953c)
Ensuite on aura l’équation variée
![{\displaystyle {\overset {.}{z}}\,\!'+2z\varphi '(z){\sqrt {1+y'^{2}}}+2\varphi (z){\frac {y'{\overset {.}{y}}\,\!'}{\sqrt {1+y'^{2}}}}=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a2d1a35c0d84a1315c2b51110e31ec704fe4779)
et par conséquent
![{\displaystyle {\begin{aligned}\operatorname {F} '(y')=&2\varphi (z){\frac {y'}{\sqrt {1+y'^{2}}}},\\\operatorname {F} '(z)=&2\varphi '(z){\sqrt {1+y'^{2}}},\quad \operatorname {F} '(z')=1.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/914e2dff33de9216259366722b98cbc813ed767f)
De là on aura
![{\displaystyle {\begin{alignedat}{2}\mathrm {Y} =&-\left({\frac {y'}{{\sqrt {1+y'^{2}}}{\sqrt {z}}}}\right)',&{\overset {\shortmid }{\mathrm {Y} }}=&{\frac {y'}{{\sqrt {1+y'^{2}}}{\sqrt {z}}}},\\\mathrm {Z} =&-{\frac {\sqrt {1+y'^{2}}}{2z^{\frac {3}{2}}}},&{\overset {\shortmid }{\mathrm {Z} }}=&0,\\(\mathrm {Y} )=&-\left[2\lambda \varphi (z){\frac {y'}{\sqrt {1+y'^{2}}}}\right]',\qquad &\left({\overset {\shortmid }{\mathrm {Y} }}\right)=&2\lambda \varphi (z){\frac {y'}{\sqrt {1+y'^{2}}}},\\(\mathrm {Z} )=&2\lambda \varphi '(z){\sqrt {1+y'^{2}}}-\lambda ',&\left({\overset {\shortmid }{\mathrm {Z} }}\right)=&\lambda .\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a57c58a06a6fc744fddb0c8388794b6a5e202ad4)
D’après ces valeurs, on aura les équations générales
![{\displaystyle \mathrm {Y+(Y)=0,\quad Z+(Z)} =0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/300e60b64524ea8b8c2e48d40e07da13c2eeb0b3)