X.
Problème V. – Déterminer si une solution
est une intégrale particulière de l’équation différentielle à trois variables
![{\displaystyle dz=pdx+qdy.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e20d5cf9b5dcdcdf25d53bb4e4deb9bb4bff3c6a)
Je suppose que cette équation soit intégrale. Cela posé, si l’on a l’équation
et que l’on veuille déterminer la variable
qui répond à
et à
on aura
![{\displaystyle z'=\varphi (x+\alpha ,y+{\text{ϐ}}).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e71c1ae1b44bd4f73ca4a892d3e0f79095ccb4b0)
Or, en supposant
constant, on a
![{\displaystyle \varphi (x+\alpha ,y+{\text{ϐ}})=\varphi (x,y+{\text{ϐ}})+\alpha {\frac {\partial \varphi (x,y+{\text{ϐ}})}{\partial x}}+{\frac {\alpha ^{2}}{1.2}}{\frac {\partial ^{2}\varphi (x,y+{\text{ϐ}})}{\partial x^{2}}}+\ldots .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db591cfda89bc5e924be8e451b7d2c0ed261ea32)
De plus, en supposant
constant,
![{\displaystyle {\begin{aligned}\varphi (x,y+{\text{ϐ}})=&\varphi (x,y)+{\text{ϐ}}{\frac {\partial \varphi (x,y)}{\partial y}}+{\frac {{\text{ϐ}}^{2}}{1.2}}{\frac {\partial ^{2}\varphi (x,y)}{\partial y^{2}}}+\ldots ,\\{\frac {\partial \varphi (x,y+{\text{ϐ}})}{\partial x}}=&{\frac {\partial \varphi (x,y)}{\partial x}}+{\text{ϐ}}{\frac {\partial ^{2}\varphi (x,y)}{\partial x\partial y}}+\ldots ,\\\ldots \ldots \ldots \ldots &\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a3376eb2e3ba55de916e27be0730cd5b4f44e32)
Donc on aura
![{\displaystyle {\begin{aligned}z'=z+\alpha {\frac {\partial z}{\partial x}}&+{\frac {\alpha ^{2}}{1.2}}{\frac {\partial ^{2}z}{\partial x^{2}}}+\ldots \\+{\text{ϐ}}{\frac {\partial z}{\partial y}}&+\alpha {\text{ϐ}}{\frac {\partial ^{2}z}{\partial x\partial y}}+\ldots \\&+{\frac {{\text{ϐ}}^{2}}{1.2}}{\frac {\partial ^{2}z}{\partial y^{2}}}+\ldots \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8185616f4f51cc0a444bea53e78b6db93d4a8f61)
L’équation
![{\displaystyle dz=pdx+qdy}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5614ae07dddf9415a5937b2e0464b7966dbbdc61)
donne
![{\displaystyle {\frac {\partial z}{\partial x}}=p,\qquad {\frac {\partial z}{\partial y}}=q.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/394a52a9cfa24dff54b2b3438e36aac430bc98fd)
Donc
![{\displaystyle {\frac {\partial ^{2}z}{\partial x^{2}}}={\frac {\partial p}{\partial x}}+{\frac {\partial p}{\partial z}}{\frac {\partial z}{\partial x}}={\frac {\partial p}{\partial x}}+p{\frac {\partial p}{\partial z}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ae34b41a341fbc596e97d8c3f798186fcae9d19)