{"id":32870,"date":"2022-04-28T13:00:17","date_gmt":"2022-04-28T13:00:17","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=32870"},"modified":"2025-04-03T23:19:40","modified_gmt":"2025-04-03T23:19:40","slug":"quest-ce-quune-equation-differentielle-ordinaire-edo","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/fr\/quest-ce-quune-equation-differentielle-ordinaire-edo\/","title":{"rendered":"Qu\u2019est-ce qu\u2019une \u00c9quation Diff\u00e9rentielle Ordinaire (EDO) ?"},"content":{"rendered":"<style>\np, ul, ol{\ntext-align: justify;\n}\nh1{\ntext-align:center;\ntext-transform: uppercase;\n}\nh2{\ntext-align:center;\ntext-transform: uppercase;\nfont-size:24pt;\n}\nh3 { \n    text-align: center;\n    text-transform: uppercase;\n    font-size: 24px !important;\n}\n<\/style>\n<h1>Qu&#8217;est-ce qu&#8217;une \u00c9quation Diff\u00e9rentielle Ordinaire (EDO) ?<\/h1>\n<p style=\"text-align:center;\" dir=\"ltr\"><em><strong>R\u00e9sum\u00e9 :<\/strong><\/br>Dans ce cours, nous explorons les \u00c9quations Diff\u00e9rentielles Ordinaires (EDO) d&#8217;ordre k, en commen\u00e7ant par leur d\u00e9finition et leur repr\u00e9sentation sous forme normale et g\u00e9n\u00e9rale. \u00c0 travers des concepts tels que la matrice jacobienne et le Th\u00e9or\u00e8me de la Fonction Implicite, nous \u00e9tablissons les bases pour comprendre les solutions de ces \u00e9quations et les propri\u00e9t\u00e9s associ\u00e9es, comme le domaine de d\u00e9finition et les solutions explicites et implicites.<\/em><\/p>\n<p style=\"text-align:center\"><strong>OBJECTIFS D&#8217;APPRENTISSAGE<\/strong><\/p>\n<p>\u00c0 la fin de ce cours, l&#8217;\u00e9tudiant sera capable de :<\/p>\n<ol>\n<li><strong>Rappeler<\/strong> la d\u00e9finition et les caract\u00e9ristiques de base d&#8217;une \u00c9quation Diff\u00e9rentielle Ordinaire (EDO).<\/li>\n<li><strong>Expliquer<\/strong> la relation entre une EDO et ses solutions possibles.<\/li>\n<\/ol>\n<p style=\"text-align:center;\" dir=\"ltr\"><strong>INDEX<\/strong><br \/>\n<a href=\"#LaEcuacionDiferencialOrdinariaDeOrdenK\"><strong>L&#8217;\u00c9quation Diff\u00e9rentielle Ordinaire (EDO) d&#8217;ordre k<\/strong><\/a><br \/>\n<a href=\"#TeoremaDeLaFuncionImplicita\">Th\u00e9or\u00e8me de la Fonction Implicite<\/a><br \/>\n<a href=\"#LaSolucionDeUnaEcuacionDiferencialOrdinaria\"><strong>La Solution d&#8217;une \u00c9quation Diff\u00e9rentielle Ordinaire<\/strong><\/a><br \/>\n<a href=\"#CuidadoConElDominioDeDefinicionDeLasSoluciones\">Attention au domaine de d\u00e9finition des solutions<\/a><br \/>\n<a href=\"#SolucionExtendidaYSolucionMaximal\">Solution \u00e9tendue et solution maximale<\/a><br \/>\n<a href=\"#SolucionExplicitaYSolucionImplicita\">Solution explicite et solution implicite<\/a>\n<\/p>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/zE29azRIKng\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/center><\/p>\n<p>Avec ce que nous avons vu jusqu&#8217;\u00e0 pr\u00e9sent, nous avons une id\u00e9e assez claire de ce qu&#8217;est une \u00e9quation diff\u00e9rentielle et des nombreuses applications qu&#8217;elle peut avoir. Nous allons maintenant nous arr\u00eater pour \u00e9tudier certaines d\u00e9finitions et propri\u00e9t\u00e9s dans le but d&#8217;\u00e9tablir une base commune solide pour poursuivre cette \u00e9tude.<\/p>\n<p><a name=\"LaEcuacionDiferencialOrdinariaDeOrdenK\"><\/a><\/p>\n<h3>L&#8217;EDO d&#8217;ordre k<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=zE29azRIKng&#038;t=163s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Une \u00c9quation Diff\u00e9rentielle Ordinaire (EDO)<\/span><\/strong><\/a> est une \u00e9quation dans laquelle interviennent une variable ind\u00e9pendante <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span>, une fonction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y(x)<\/span><\/span>, et certaines de ses d\u00e9riv\u00e9es ordinaires. Les d\u00e9riv\u00e9es ordinaires du premier ordre de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y(x)<\/span><\/span> sont not\u00e9es \u00e0 l\u2019aide de symboles tels que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\frac{dy(x)}{dx}<\/span><\/span> ou <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y&#039;(x)<\/span><\/span>, celles du second ordre comme <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\frac{d^2y(x)}{dx^2}<\/span><\/span> ou <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y&#039;&#039;(x)<\/span><\/span>, et en g\u00e9n\u00e9ral, d&#8217;ordre <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>, comme <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\frac{d^ny(x)}{dx^n}<\/span><\/span> ou <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y^{(n)}(x)<\/span><\/span>. Le plus grand des entiers <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> tels que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y^{(k)}(x)<\/span><\/span> appara\u00eet dans l&#8217;\u00e9quation est ce que nous appelons l\u2019<strong>ordre de l\u2019\u00e9quation<\/strong>. Ainsi, la <strong>forme g\u00e9n\u00e9rale d&#8217;une EDO d&#8217;ordre <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span><\/strong> est :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F\\left(x,y(x),y&#039;(x), \\cdots, y^{(k)}(x)\\right)=0.<\/span>\n<p>On dit qu&#8217;une EDO d&#8217;ordre <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> est sous <strong>forme normale<\/strong> si elle est exprim\u00e9e en isolant <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y^{(k)}(x)<\/span><\/span> de l&#8217;\u00e9quation pr\u00e9c\u00e9dente, c&#8217;est-\u00e0-dire :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y^{(k)}(x) = f\\left(x,y(x),y&#039;(x), \\cdots, y^{(k-1)}(x)\\right).<\/span>\n<p>En g\u00e9n\u00e9ral, la fonction <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> est une fonction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R} \\longrightarrow \\mathbb{R}^n,<\/span><\/span> de sorte que cette fonction et toutes ses d\u00e9riv\u00e9es \u00e9valu\u00e9es en un point <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in\\mathbb{R}<\/span><\/span> sont des vecteurs de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span><\/span>. En tenant compte de cela, on constate que, puisque la fonction <span class=\"katex-eq\" data-katex-display=\"false\">F<\/span> qui d\u00e9crit l&#8217;EDO d&#8217;ordre <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> a <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">1+(k+1)<\/span><\/span> variables, on a <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\text{Dom}(F)\\subset \\mathbb{R}^{1+n(k+1)}<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\text{Rec}(F)\\subset \\mathbb{R}<\/span><\/span>; et de mani\u00e8re analogue, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\text{Dom}(f) = \\mathbb{R}^{1+nk}<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\text{Rec}(f)\\subset \\mathbb{R}^n<\/span><\/span>.<\/p>\n<p>Le passage de l\u2019expression g\u00e9n\u00e9rale d\u2019une EDO d\u2019ordre <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> \u00e0 sa forme normale est possible gr\u00e2ce au <strong>Th\u00e9or\u00e8me de la Fonction Implicite.<\/strong><\/p>\n<p><a name=\"TeoremaDeLaFuncionImplicita\"><\/a><\/p>\n<h4>Th\u00e9or\u00e8me de la Fonction Implicite<\/h4>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=zE29azRIKng&#038;t=887s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Soit <span class=\"katex-eq\" data-katex-display=\"false\">F<\/span> une fonction de classe <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathcal{C}^1<\/span><\/span> sur un ensemble ouvert <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">U \\subset \\mathbb{R}^n<\/span><\/span><\/span><\/strong><\/a> \u00e0 valeurs r\u00e9elles. Et soit <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(a_1,\\cdots, a_n) \\in U<\/span><\/span> tel que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F(a_1,\\cdots, a_n) = 0<\/span><\/span> et<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{\\partial F(a_1,\\cdots, a_n)}{\\partial x_n} \\neq 0<\/span>\n<p>Alors il existe un voisinage <span class=\"katex-eq\" data-katex-display=\"false\">V<\/span> de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(a_1, \\cdots, a_{n-1}) \\in \\mathbb{R}^{n-1}<\/span><\/span> et une fonction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\varphi:V \\longrightarrow \\mathbb{R}<\/span><\/span> telle que :<\/p>\n<ol>\n<li type=\"i\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">V \\times \\varphi(V) \\subset U<\/span><\/span><\/li>\n<li type=\"i\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F(x_1,\\cdots,x_{n-1},x_n) = 0 \\leftrightarrow x_n = \\varphi(x_1,\\cdots, x_{n-1})<\/span><\/span><\/li>\n<li type=\"i\"><span class=\"katex-eq\" data-katex-display=\"false\">\\varphi<\/span> est diff\u00e9rentiable et\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\\dfrac{\\partial \\varphi (a_1,\\cdots, a_{n-1})}{\\partial x_i} = - \\dfrac{ \\dfrac{\\partial F (a_1,\\cdots, a_n)}{\\partial x_i} }{ \\dfrac{\\partial F (a_1,\\cdots, a_n)}{\\partial x_n} }<\/span>\n<\/li>\n<\/ol>\n<h4>D\u00e9monstration du Th\u00e9or\u00e8me de la Fonction Implicite<\/h4>\n<h5>D\u00e9veloppement \u00e0 partir de la matrice Jacobienne<\/h5>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=zE29azRIKng&#038;t=1101s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Soit <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi(x_1,\\cdots,x_{n-1}, x_n) = (x_1,\\cdots,x_{n-1}, F(x_1,\\cdots, x_n)).<\/span><\/span><\/span><\/strong><\/a> Si l&#8217;on calcule sa matrice Jacobienne, qui est montr\u00e9e ci-dessous :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left( \\dfrac{\\partial \\psi(x_1,\\cdots, x_n)}{\\partial(x_1,\\cdots, x_n)} \\right) = \\left( \\begin{array}{cccc}\n\n1 &amp; 0 &amp;  \\cdots &amp; 0 \\\\\n\n0 &amp; 1 &amp;  \\cdots &amp; \\vdots \\\\\n\n\\vdots &amp;\\vdots &amp; \\ddots  &amp; \\vdots  \\\\\n\n\\displaystyle \\dfrac{\\partial F(x_1, \\cdots, x_n)}{\\partial x_1} &amp; \\dfrac{\\partial F(x_1, \\cdots, x_n)}{\\partial x_2} &amp; \\cdots  &amp; \\dfrac{\\partial F(x_1, \\cdots, x_n)}{\\partial x_n}\n\n\\end{array}\\right), <\/span>\n<p>on verra que son d\u00e9terminant est diff\u00e9rent de z\u00e9ro en <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(a_1,\\cdots, a_n)<\/span><\/span>, pr\u00e9cis\u00e9ment parce que, comme \u00e9tabli au d\u00e9but, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\partial F(a_1,\\cdots, a_n)\/\\partial x_n \\neq 0.<\/span><\/span> \u00c0 partir de cela, on peut dire que <span class=\"katex-eq\" data-katex-display=\"false\">\\psi<\/span> admet une inverse sur un ensemble ouvert <span class=\"katex-eq\" data-katex-display=\"false\">W<\/span> contenant <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(a_1,\\cdots, a_n).<\/span><\/span><\/p>\n<h5>D\u00e9veloppement de la solution<\/h5>\n<p>Consid\u00e9rons maintenant un ensemble<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\tilde{V}=\\psi(W)\\ni \\psi(a_1,\\cdots,a_{n}) = (a_1,\\cdots,a_{n-1},F(a_1,\\cdots,a_{n}))=(a_1,\\cdots,a_{n-1},0).<\/span>\n<p>\u00c0 partir de cela, on peut d\u00e9finir un autre ensemble<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">V=\\{(x_1,\\cdots,x_{n-1}) \\;|\\; (x_1,\\cdots,x_{n-1},0)\\in \\tilde{V}\\}\\ni (a_1,\\cdots,a_{n-1})<\/span>\n<p>L&#8217;ensemble <span class=\"katex-eq\" data-katex-display=\"false\">V<\/span> est, par cons\u00e9quent, un ouvert contenant <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(a_1,\\cdots,a_{n-1})\\in\\mathbb{R}^{n-1}.<\/span><\/span><\/p>\n<p>De plus, comme <span class=\"katex-eq\" data-katex-display=\"false\">\\psi<\/span> est inversible (dans <span class=\"katex-eq\" data-katex-display=\"false\">W<\/span>), il existe un unique <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(y_1,\\cdots,y_n)\\in W<\/span><\/span> tel que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi(y_1,\\cdots,y_n) = (x_1,\\cdots,x_{n-1},0).<\/span><\/span> Cela signifie que :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl} y_1 &amp;= x_1 \\\\ \\\\ \\vdots &amp; \\vdots \\\\ \\\\ y_{n-1} &amp;= x_{n-1} \\\\ \\\\ F(x_1,\\cdots,x_{n-1},y_n) &amp;= 0 \\end{array}<\/span>\n<p>Ainsi, on peut d\u00e9finir <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\varphi(x_1,\\cdots,x_{n-1}) = y_n<\/span><\/span>, de sorte que :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi^{-1}(x_1,\\cdots,x_{n-1},0) = (x_1,\\cdots,x_{n-1},\\varphi(x_1,\\cdots,x_{n-1}))<\/span>\n<p>et<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F(x_1,\\cdots,x_{n-1},\\varphi(x_1,\\cdots,x_{n-1})) = 0<\/span>\n<p>\u00c0 partir de cela, on a que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\varphi(V)\\ni a_n,<\/span><\/span> et par cons\u00e9quent <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">V\\times\\varphi(V) \\subset U,<\/span><\/span> et en outre :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F(x_1,\\cdots,x_{n-1},x_n) = 0 \\leftrightarrow x_n = \\varphi(x_1,\\cdots,x_{n-1})<\/span>\n<h5>D\u00e9rivabilit\u00e9<\/h5>\n<p>Et finalement, la diff\u00e9rentiabilit\u00e9 de <span class=\"katex-eq\" data-katex-display=\"false\">\\psi<\/span> entra\u00eene la diff\u00e9rentiabilit\u00e9 de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi^{-1}<\/span><\/span>, ce qui entra\u00eene \u00e0 son tour celle de <span class=\"katex-eq\" data-katex-display=\"false\">\\varphi<\/span> sur <span class=\"katex-eq\" data-katex-display=\"false\">V<\/span>. En tenant compte de cela, on peut d\u00e9finir une fonction <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> \u00e0 travers la relation :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g(x_1, \\cdots,x_{n-1}) = F(x_1,\\cdots,x_{n-1},\\varphi(x_1,\\cdots,x_{n-1})) = 0<\/span>\n<p>Et ensuite, en utilisant la r\u00e8gle de la cha\u00eene, on a :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{\\partial g}{\\partial x_i} = \\frac{\\partial F}{\\partial x_i} + \\frac{\\partial F}{\\partial x_n}\\frac{\\partial \\varphi }{\\partial x_i} = 0,<\/span>\n<p>o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">i=1,\\cdots, n-1.<\/span><\/span> C\u2019est \u00e0 partir de cette derni\u00e8re \u00e9quation que l\u2019on obtient :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\dfrac{\\partial \\varphi(a_1,\\cdots,a_{n-1})}{\\partial x_i} = - \\dfrac{\\dfrac{\\partial F(a_1,\\cdots,a_{n})}{\\partial x_i}}{\\dfrac{\\partial F(a_1,\\cdots,a_{n})}{\\partial x_n}}<\/span>\n<p>Et avec cela, on conclut tout ce que l\u2019on voulait d\u00e9montrer \u25a0<\/p>\n<p><a name=\"LaSolucionDeUnaEcuacionDiferencialOrdinaria\"><\/a><\/p>\n<h3>La solution d&#8217;une \u00e9quation diff\u00e9rentielle ordinaire<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=zE29azRIKng&#038;t=2249s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Consid\u00e9rons une EDO exprim\u00e9e sous forme normale<\/span><\/strong><\/a><\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y^{(n)} = f(x,y(x),y^\\prime(x),\\cdots,y^{(n-1)(x)})<\/span>\n<p>Alors, une fonction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\varphi : I_\\phi \\longmapsto \\mathbb{R}^n,<\/span><\/span> o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_\\phi<\/span><\/span> est un intervalle de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R},<\/span><\/span> est dite <strong>une solution de l&#8217;EDO<\/strong> si :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left(\\forall x \\in I_\\phi \\right) \\left(\\varphi^{(n)}(x) = f(x,\\varphi(x),\\varphi^\\prime(x),\\cdots,\\varphi^{(n-1)(x)}\\right)<\/span>\n<p><a name=\"CuidadoConElDominioDeDefinicionDeLasSoluciones\"><\/a><\/p>\n<h4>Attention au domaine de d\u00e9finition des solutions<\/h4>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=zE29azRIKng&#038;t=2387s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">\u00c0 ce stade, il est n\u00e9cessaire de souligner<\/span><\/strong><\/a> l&#8217;importance de d\u00e9clarer explicitement le domaine de la solution de l&#8217;\u00e9quation diff\u00e9rentielle. Par exemple, le domaine de la fonction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi<\/span><\/span> mentionn\u00e9e dans le paragraphe pr\u00e9c\u00e9dent est l&#8217;intervalle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_\\phi.<\/span><\/span> Cela est important car une erreur fr\u00e9quente dans le traitement des \u00e9quations diff\u00e9rentielles est de consid\u00e9rer comme \u00e9gales deux solutions <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi_1<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi_2<\/span><\/span> uniquement parce que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left(\\forall x \\in I_{\\phi_1}\\cap I_{\\phi_2}\\right)\\left(\\phi_1(x) = \\phi_2(x)\\right),<\/span><\/span> m\u00eame si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{\\phi_1}\\neq I_{\\phi_2}.<\/span><\/span> Pour expliquer ce point, examinons l&#8217;\u00e9quation diff\u00e9rentielle :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y^\\prime = -y^2.<\/span>\n<p>Une solution possible de cette EDO est la fonction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_1 : ]0,+\\infty[ \\longrightarrow \\mathbb{R}^+\\setminus\\{0\\}<\/span><\/span> d\u00e9finie par <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_1(x)=1\/x,<\/span><\/span> car <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_1^{\\prime} = -1\/x^2 = -\\psi_1^2<\/span><\/span> pour tout <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in]0,+\\infty[.<\/span><\/span> Mais en jouant un peu avec l&#8217;alg\u00e8bre, on peut obtenir une autre solution compl\u00e8tement diff\u00e9rente si l&#8217;on ne pr\u00eate pas attention aux d\u00e9tails. Par exemple, il est clair que :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{1}{x} = \\frac{1}{1 - (1-x)},<\/span>\n<p>et le c\u00f4t\u00e9 droit de cette \u00e9galit\u00e9 est le r\u00e9sultat de la s\u00e9rie g\u00e9om\u00e9trique :<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle \\sum_{n=0}^{+\\infty} (1-x)^n = \\frac{1}{1 - (1-x)}<\/span>\n<p>Ainsi, un \u0153il peu entra\u00een\u00e9 \u00e0 ces arts arcaniques pourrait \u00eatre tent\u00e9 de croire que les fonctions <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_1<\/span><\/span><br \/>\n et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_2 = \\sum_{n=0}^{+\\infty} (1-x)^n <\/span><\/span> nous donnent la m\u00eame solution pour l&#8217;\u00e9quation diff\u00e9rentielle initiale, car elles co\u00efncident effectivement dans leurs r\u00e9sultats ; cependant, cela passerait \u00e0 c\u00f4t\u00e9 du fait que cette s\u00e9rie g\u00e9om\u00e9trique n&#8217;est valable que lorsque <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|1-x| \\lt 1<\/span><\/span>, c\u2019est-\u00e0-dire lorsque <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in]0,2[)<\/span><\/span>. Mais il y a plus : puisque <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">]0,2[\\subset]0,+\\infty[<\/span><\/span>, on a \u00e9galement que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_1<\/span><\/span> prolonge <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_2<\/span><\/span>, car l\u00e0 o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_2<\/span><\/span> est valide, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_1<\/span><\/span> l&#8217;est aussi \u2014 et au-del\u00e0.<\/p>\n<p><a name=\"SolucionExtendidaYSolucionMaximal\"><\/a><\/p>\n<h4>Solution \u00e9tendue et solution maximale<\/h4>\n<p>Consid\u00e9rons deux fonctions <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi_1<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi_2<\/span><\/span> d\u00e9finies sur les intervalles <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{\\phi_1}<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{\\phi_2},<\/span><\/span> respectivement, qui sont des solutions d&#8217;une \u00e9quation diff\u00e9rentielle. Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{\\phi_1}\\subset I_{\\phi_2},<\/span><\/span> on dit alors que la solution <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi_2<\/span><\/span> \u00e9tend la solution <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi_1,<\/span><\/span> ou que la solution <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi_2<\/span><\/span> est plus g\u00e9n\u00e9rale que la solution <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi_1.<\/span><\/span> Une solution <span class=\"katex-eq\" data-katex-display=\"false\">\\phi<\/span> est dite \u00abmaximale\u00bb s\u2019il n\u2019existe pas d\u2019autre solution qui l\u2019\u00e9tend de mani\u00e8re non triviale.<\/p>\n<p><a name=\"SolucionExplicitaYSolucionImplicita\"><\/a><\/p>\n<h4>Solution explicite et solution implicite<\/h4>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=zE29azRIKng&#038;t=2649s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Une fonction<\/span><\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">\\phi<\/span> est consid\u00e9r\u00e9e comme solution de l&#8217;EDO d&#8217;ordre <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> (\u00e9crite sous forme normale)<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y^{(n)}(x)=f(x,y(x),y^\\prime(x),\\cdots,y^{(n-1)}(x)),<\/span>\n<p> sur un intervalle <span class=\"katex-eq\" data-katex-display=\"false\">I<\/span> si<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(\\forall x\\in I)\\left(\\phi^{n}(x) = f(x,\\phi(x),\\phi^\\prime(x),\\cdots,\\phi^{(n-1)}(x))\\right)<\/span>\n<p>Ce que nous avions d\u00e9j\u00e0 vu plusieurs paragraphes plus t\u00f4t est ce que l&#8217;on appelle une <strong>solution explicite de l&#8217;\u00e9quation diff\u00e9rentielle sur l&#8217;intervalle <span class=\"katex-eq\" data-katex-display=\"false\">I.<\/span><\/strong> Comme le nom l&#8217;indique, il existe \u00e9galement une forme implicite pour d\u00e9finir les solutions. On dit qu\u2019une relation <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Phi(x,y)=0<\/span><\/span> est une <strong>solution implicite de l&#8217;\u00e9quation diff\u00e9rentielle sur <span class=\"katex-eq\" data-katex-display=\"false\">I<\/span><\/strong> si elle d\u00e9finit deux ou plusieurs solutions implicites sur <span class=\"katex-eq\" data-katex-display=\"false\">I.<\/span>\n<h3>Conclusion<\/h3>\n<p>Dans ce cours, nous avons d\u00e9compos\u00e9 la notion d&#8217;\u00e9quation diff\u00e9rentielle ordinaire avec une approche rigoureuse mais accessible, en posant les fondements formels qui nous permettent non seulement de reconna\u00eetre une EDO, mais aussi de comprendre la logique qui sous-tend ses solutions. Gr\u00e2ce au Th\u00e9or\u00e8me de la Fonction Implicite, il a \u00e9t\u00e9 possible de justifier clairement la transition entre sa forme g\u00e9n\u00e9rale et sa forme normale, ce qui constitue une capacit\u00e9 technique essentielle pour aborder des probl\u00e8mes concrets.<\/p>\n<p>Nous avons \u00e9galement distingu\u00e9 avec pr\u00e9cision les diff\u00e9rentes mani\u00e8res dont une solution peut \u00eatre comprise : comme solution explicite ou implicite, \u00e9tendue ou maximale, et soulign\u00e9 l&#8217;importance \u2014 souvent sous-estim\u00e9e \u2014 de bien d\u00e9clarer son domaine. Ces distinctions ne sont pas seulement formelles : elles sont op\u00e9ratoires. Les ignorer peut, comme nous l&#8217;avons vu, entra\u00eener de graves erreurs conceptuelles dans l\u2019interpr\u00e9tation des r\u00e9sultats obtenus.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Qu&#8217;est-ce qu&#8217;une \u00c9quation Diff\u00e9rentielle Ordinaire (EDO) ? R\u00e9sum\u00e9 :Dans ce cours, nous explorons les \u00c9quations Diff\u00e9rentielles Ordinaires (EDO) d&#8217;ordre k, en commen\u00e7ant par leur d\u00e9finition et leur repr\u00e9sentation sous forme normale et g\u00e9n\u00e9rale. \u00c0 travers des concepts tels que la matrice jacobienne et le Th\u00e9or\u00e8me de la Fonction Implicite, nous \u00e9tablissons les bases pour comprendre [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":32838,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":18,"footnotes":""},"categories":[1162,569,11],"tags":[],"class_list":["post-32870","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-equations-differentielles-ordinaires","category-mathematiques","category-sin-categoria-fr"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Qu\u2019est-ce qu\u2019une \u00c9quation Diff\u00e9rentielle Ordinaire (EDO) ? 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