{"id":32846,"date":"2022-04-28T13:00:21","date_gmt":"2022-04-28T13:00:21","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=32846"},"modified":"2025-04-03T22:49:36","modified_gmt":"2025-04-03T22:49:36","slug":"o-que-e-uma-equacao-diferencial-ordinaria-edo","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/pt\/o-que-e-uma-equacao-diferencial-ordinaria-edo\/","title":{"rendered":"O que \u00e9 uma Equa\u00e7\u00e3o Diferencial Ordin\u00e1ria (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>O que \u00e9 uma Equa\u00e7\u00e3o Diferencial Ordin\u00e1ria (EDO)?<\/h1>\n<p style=\"text-align:center;\" dir=\"ltr\"><em><strong>Resumo:<\/strong><\/br>Nesta aula, exploram-se as Equa\u00e7\u00f5es Diferenciais Ordin\u00e1rias (EDO) de ordem k, come\u00e7ando com sua defini\u00e7\u00e3o e sua representa\u00e7\u00e3o de maneira normal e geral. Atrav\u00e9s de conceitos como a matriz Jacobiana e o Teorema da Fun\u00e7\u00e3o Impl\u00edcita, constroem-se as bases para compreender as solu\u00e7\u00f5es dessas equa\u00e7\u00f5es e as propriedades associadas, como o dom\u00ednio de defini\u00e7\u00e3o e as solu\u00e7\u00f5es expl\u00edcitas e impl\u00edcitas.<\/em><\/p>\n<p style=\"text-align:center\"><strong>OBJETIVOS DE APRENDIZADO<\/strong><\/p>\n<p>Ao final desta aula, o estudante ser\u00e1 capaz de:<\/p>\n<ol>\n<li><strong>Recordar<\/strong> a defini\u00e7\u00e3o e caracter\u00edsticas b\u00e1sicas de uma Equa\u00e7\u00e3o Diferencial Ordin\u00e1ria (EDO).<\/li>\n<li><strong>Explicar<\/strong> a rela\u00e7\u00e3o entre uma EDO e suas poss\u00edveis solu\u00e7\u00f5es.<\/li>\n<\/ol>\n<p style=\"text-align:center;\" dir=\"ltr\"><strong>\u00cdNDICE<\/strong><br \/>\n<a href=\"#LaEcuacionDiferencialOrdinariaDeOrdenK\"><strong>A Equa\u00e7\u00e3o Diferencial Ordin\u00e1ria (EDO) de Ordem k<\/strong><\/a><br \/>\n<a href=\"#TeoremaDeLaFuncionImplicita\">Teorema da Fun\u00e7\u00e3o Impl\u00edcita<\/a><br \/>\n<a href=\"#LaSolucionDeUnaEcuacionDiferencialOrdinaria\"><strong>A Solu\u00e7\u00e3o de uma Equa\u00e7\u00e3o Diferencial Ordin\u00e1ria<\/strong><\/a><br \/>\n<a href=\"#CuidadoConElDominioDeDefinicionDeLasSoluciones\">Cuidado com o dom\u00ednio de defini\u00e7\u00e3o das solu\u00e7\u00f5es<\/a><br \/>\n<a href=\"#SolucionExtendidaYSolucionMaximal\">Solu\u00e7\u00e3o estendida e solu\u00e7\u00e3o maximal<\/a><br \/>\n<a href=\"#SolucionExplicitaYSolucionImplicita\">Solu\u00e7\u00e3o expl\u00edcita e solu\u00e7\u00e3o impl\u00edcita<\/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>Com o que vimos at\u00e9 agora, temos uma ideia bastante clara do que \u00e9 uma equa\u00e7\u00e3o diferencial e das m\u00faltiplas aplica\u00e7\u00f5es que podem ter. Vamos agora nos deter para estudar algumas defini\u00e7\u00f5es e propriedades com o objetivo de estabelecer uma base comum s\u00f3lida para continuar este estudo.<\/p>\n<p><a name=\"LaEcuacionDiferencialOrdinariaDeOrdenK\"><\/a><\/p>\n<h3>A EDO de Ordem 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;\">Uma Equa\u00e7\u00e3o Diferencial Ordin\u00e1ria (EDO)<\/span><\/strong><\/a> \u00e9 uma equa\u00e7\u00e3o na qual est\u00e3o envolvidas uma vari\u00e1vel independente <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span>, uma fun\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y(x)<\/span><\/span>, e algumas de suas derivadas ordin\u00e1rias. As derivadas ordin\u00e1rias de primeira ordem de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y(x)<\/span><\/span> s\u00e3o denotadas por s\u00edmbolos como <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>, as de segunda ordem como <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>, e em geral, de ordem <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>, como <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>. O supremo dos valores <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> tais que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y^{(k)}(x)<\/span><\/span> aparece na equa\u00e7\u00e3o \u00e9 o que chamamos de <strong>Ordem da Equa\u00e7\u00e3o<\/strong>. Desse modo, a <strong>Forma Geral de uma EDO de ordem <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span><\/strong> \u00e9:<\/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>Diz-se que uma EDO de ordem <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> est\u00e1 na <strong>forma normal<\/strong> se for expressa isolando <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y^{(k)}(x)<\/span><\/span> da equa\u00e7\u00e3o anterior, ou seja:<\/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>Em geral, a fun\u00e7\u00e3o <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> \u00e9 uma fun\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R} \\longrightarrow \\mathbb{R}^n,<\/span><\/span> de modo que essa e todas as suas derivadas avaliadas em algum ponto <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in\\mathbb{R}<\/span><\/span> s\u00e3o vetores de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span><\/span>. Com isso em considera\u00e7\u00e3o, constata-se que, dado que a fun\u00e7\u00e3o <span class=\"katex-eq\" data-katex-display=\"false\">F<\/span> que descreve a EDO de ordem <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> possui <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">1+(k+1)<\/span><\/span> vari\u00e1veis, tem-se que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\text{Dom}(F)\\subset \\mathbb{R}^{1+n(k+1)}<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\text{Rec}(F)\\subset \\mathbb{R}<\/span><\/span>; e de forma an\u00e1loga, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\text{Dom}(f) = \\mathbb{R}^{1+nk}<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\text{Rec}(f)\\subset \\mathbb{R}^n<\/span><\/span>.<\/p>\n<p>A passagem da express\u00e3o Geral de uma EDO de ordem <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> para sua Forma Normal \u00e9 poss\u00edvel gra\u00e7as ao <strong>Teorema da Fun\u00e7\u00e3o Impl\u00edcita.<\/strong><\/p>\n<p><a name=\"TeoremaDeLaFuncionImplicita\"><\/a><\/p>\n<h4>Teorema da Fun\u00e7\u00e3o Impl\u00edcita<\/h4>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=zE29azRIKng&#038;t=887s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Seja <span class=\"katex-eq\" data-katex-display=\"false\">F<\/span> uma fun\u00e7\u00e3o de classe <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathcal{C}^1<\/span><\/span> sobre um conjunto aberto <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">U \\subset \\mathbb{R}^n<\/span><\/span><\/span><\/strong><\/a> com valores reais. E seja <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(a_1,\\cdots, a_n) \\in U<\/span><\/span> tal que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F(a_1,\\cdots, a_n) = 0<\/span><\/span> e<\/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>Ent\u00e3o existe uma vizinhan\u00e7a <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> e uma fun\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\varphi:V \\longrightarrow \\mathbb{R}<\/span><\/span> tal 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> \u00e9 diferenci\u00e1vel e\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>Demonstra\u00e7\u00e3o do Teorema da Fun\u00e7\u00e3o Impl\u00edcita<\/h4>\n<h5>Desenvolvimento a partir da matriz Jacobiana<\/h5>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=zE29azRIKng&#038;t=1101s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Seja <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> Se calcularmos sua matriz Jacobiana, que \u00e9 mostrada a seguir:<\/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>veremos que seu determinante \u00e9 diferente de zero em <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(a_1,\\cdots, a_n)<\/span><\/span>, precisamente porque, como se estabeleceu no in\u00edcio, <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> A partir disso, podemos dizer que <span class=\"katex-eq\" data-katex-display=\"false\">\\psi<\/span> tem uma inversa sobre um conjunto aberto <span class=\"katex-eq\" data-katex-display=\"false\">W<\/span> que cont\u00e9m <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(a_1,\\cdots, a_n).<\/span><\/span><\/p>\n<h5>Desenvolvimento da Solu\u00e7\u00e3o<\/h5>\n<p>Agora, consideremos um conjunto<\/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>A partir disso, podemos definir outro conjunto<\/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>O conjunto <span class=\"katex-eq\" data-katex-display=\"false\">V<\/span> \u00e9, em consequ\u00eancia, um aberto que cont\u00e9m <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>Al\u00e9m disso, como <span class=\"katex-eq\" data-katex-display=\"false\">\\psi<\/span> tem inversa (em <span class=\"katex-eq\" data-katex-display=\"false\">W<\/span>), existe um \u00fanico <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(y_1,\\cdots,y_n)\\in W<\/span><\/span> tal 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> Isso significa 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>Assim, podemos definir <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\varphi(x_1,\\cdots,x_{n-1}) = y_n<\/span><\/span>, de modo 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>e<\/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>A partir disso, temos que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\varphi(V)\\ni a_n,<\/span><\/span> e em consequ\u00eancia <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">V\\times\\varphi(V) \\subset U,<\/span><\/span> e al\u00e9m disso:<\/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>Diferenciabilidade<\/h5>\n<p>E finalmente, a diferenciabilidade de <span class=\"katex-eq\" data-katex-display=\"false\">\\psi<\/span> conduz \u00e0 diferenciabilidade de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi^{-1}<\/span><\/span>, que por sua vez conduz \u00e0 diferenciabilidade de <span class=\"katex-eq\" data-katex-display=\"false\">\\varphi<\/span> sobre <span class=\"katex-eq\" data-katex-display=\"false\">V<\/span>. Tendo isso em conta, podemos definir uma fun\u00e7\u00e3o <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> atrav\u00e9s da rela\u00e7\u00e3o:<\/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>E ent\u00e3o, usando a regra da cadeia, temos:<\/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>onde <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">i=1,\\cdots, n-1.<\/span><\/span> \u00c9 a partir dessa \u00faltima equa\u00e7\u00e3o que se obt\u00e9m:<\/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>E com isso conclui-se tudo o que se queria demonstrar \u25a0<\/p>\n<p><a name=\"LaSolucionDeUnaEcuacionDiferencialOrdinaria\"><\/a><\/p>\n<h3>A solu\u00e7\u00e3o de uma equa\u00e7\u00e3o diferencial ordin\u00e1ria<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=zE29azRIKng&#038;t=2249s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Consideremos uma EDO expressa em forma normal<\/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>Ent\u00e3o, uma fun\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\varphi : I_\\phi \\longmapsto \\mathbb{R}^n,<\/span><\/span> onde <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_\\phi<\/span><\/span> \u00e9 um intervalo de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R},<\/span><\/span> diz-se que \u00e9 <strong>uma solu\u00e7\u00e3o da EDO<\/strong> se:<\/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>Cuidado com o dom\u00ednio de defini\u00e7\u00e3o das solu\u00e7\u00f5es<\/h4>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=zE29azRIKng&#038;t=2387s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Neste ponto, \u00e9 necess\u00e1rio enfatizar<\/span><\/strong><\/a> a import\u00e2ncia de declarar explicitamente o dom\u00ednio da solu\u00e7\u00e3o da equa\u00e7\u00e3o diferencial. Por exemplo, o dom\u00ednio da fun\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi<\/span><\/span> da qual falamos no par\u00e1grafo anterior \u00e9 o intervalo <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_\\phi.<\/span><\/span> Isso \u00e9 importante porque um erro comum ao trabalhar com equa\u00e7\u00f5es diferenciais vem de considerar iguais duas solu\u00e7\u00f5es <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi_1<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi_2<\/span><\/span> apenas porque <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> apesar de que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{\\phi_1}\\neq I_{\\phi_2}.<\/span><\/span> Para explicar esse ponto, examinemos a equa\u00e7\u00e3o diferencial:<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y^\\prime = -y^2.<\/span>\n<p>Uma poss\u00edvel solu\u00e7\u00e3o para essa EDO \u00e9 a fun\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_1 : ]0,+\\infty[ \\longrightarrow \\mathbb{R}^+\\setminus\\{0\\}<\/span><\/span> definida por <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_1(x)=1\/x,<\/span><\/span> porque <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_1^{\\prime} = -1\/x^2 = -\\psi_1^2<\/span><\/span> para qualquer <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in]0,+\\infty[.<\/span><\/span> Mas fazendo um pouco de manipula\u00e7\u00e3o alg\u00e9brica, podemos passar desta para outra solu\u00e7\u00e3o completamente diferente se n\u00e3o prestarmos aten\u00e7\u00e3o aos detalhes. Por exemplo, \u00e9 claro 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>e o lado direito dessa igualdade \u00e9 o resultado da s\u00e9rie geom\u00e9trica:<\/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>De modo que um olhar pouco treinado nessas artes arcanas se aventuraria a pensar que as fun\u00e7\u00f5es <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_1<\/span><\/span><br \/>\n e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_2 = \\sum_{n=0}^{+\\infty} (1-x)^n <\/span><\/span> nos oferecem a mesma solu\u00e7\u00e3o para a equa\u00e7\u00e3o diferencial apresentada no in\u00edcio, porque de fato coincidem em seus resultados; no entanto, ter\u00e1 passado despercebido que essa s\u00e9rie geom\u00e9trica s\u00f3 \u00e9 v\u00e1lida quando <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|1-x| \\lt 1<\/span><\/span>, ou seja, quando <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in]0,2[)<\/span><\/span>. Mas h\u00e1 mais: dado que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">]0,2[\\subset]0,+\\infty[<\/span><\/span>, tamb\u00e9m se tem que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_1<\/span><\/span> estende a <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_2<\/span><\/span> porque onde <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_2<\/span><\/span> \u00e9 v\u00e1lida, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\psi_1<\/span><\/span> tamb\u00e9m \u00e9 \u2014 e ainda al\u00e9m.<\/p>\n<p><a name=\"SolucionExtendidaYSolucionMaximal\"><\/a><\/p>\n<h4>Solu\u00e7\u00e3o estendida e solu\u00e7\u00e3o maximal<\/h4>\n<p>Consideremos duas fun\u00e7\u00f5es <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi_1<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi_2<\/span><\/span> definidas sobre os intervalos <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{\\phi_1}<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{\\phi_2},<\/span><\/span> respectivamente, que s\u00e3o solu\u00e7\u00f5es de uma equa\u00e7\u00e3o diferencial. Se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{\\phi_1}\\subset I_{\\phi_2},<\/span><\/span> ent\u00e3o diz-se que a solu\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi_2<\/span><\/span> estende a solu\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi_1,<\/span><\/span> ou que a solu\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi_2<\/span><\/span> \u00e9 mais geral que a solu\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\phi_1.<\/span><\/span> Uma solu\u00e7\u00e3o <span class=\"katex-eq\" data-katex-display=\"false\">\\phi<\/span> \u00e9 denominada \u00abmaximal\u00bb se n\u00e3o existe outra solu\u00e7\u00e3o que a estenda de forma n\u00e3o trivial.<\/p>\n<p><a name=\"SolucionExplicitaYSolucionImplicita\"><\/a><\/p>\n<h4>Solu\u00e7\u00e3o expl\u00edcita e solu\u00e7\u00e3o impl\u00edcita<\/h4>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=zE29azRIKng&#038;t=2649s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Uma fun\u00e7\u00e3o<\/span><\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">\\phi<\/span> \u00e9 considerada solu\u00e7\u00e3o da EDO de ordem <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> (escrita em forma normal)<\/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> dentro de um intervalo <span class=\"katex-eq\" data-katex-display=\"false\">I<\/span> se<\/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>O que j\u00e1 hav\u00edamos revisado h\u00e1 v\u00e1rios par\u00e1grafos \u00e9 o que se conhece como <strong>Solu\u00e7\u00e3o Expl\u00edcita da Equa\u00e7\u00e3o Diferencial no intervalo <span class=\"katex-eq\" data-katex-display=\"false\">I.<\/span><\/strong> Tal como o nome sugere, existe tamb\u00e9m uma forma impl\u00edcita de definir as solu\u00e7\u00f5es. Diz-se que uma rela\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Phi(x,y)=0<\/span><\/span> \u00e9 <strong>Solu\u00e7\u00e3o Impl\u00edcita da Equa\u00e7\u00e3o Diferencial em <span class=\"katex-eq\" data-katex-display=\"false\">I<\/span><\/strong> se define duas ou mais solu\u00e7\u00f5es impl\u00edcitas em <span class=\"katex-eq\" data-katex-display=\"false\">I.<\/span>\n<h3>Conclus\u00e3o<\/h3>\n<p>Nesta aula, decompusemos a no\u00e7\u00e3o de equa\u00e7\u00e3o diferencial ordin\u00e1ria com um olhar rigoroso, por\u00e9m acess\u00edvel, estabelecendo os fundamentos formais que nos permitem n\u00e3o apenas reconhecer uma EDO, mas tamb\u00e9m entender a l\u00f3gica por tr\u00e1s de suas solu\u00e7\u00f5es. Gra\u00e7as ao Teorema da Fun\u00e7\u00e3o Impl\u00edcita, foi poss\u00edvel justificar com clareza a transi\u00e7\u00e3o entre sua forma geral e sua forma normal, o que se traduz em uma capacidade t\u00e9cnica crucial para abordar problemas concretos.<\/p>\n<p>Al\u00e9m disso, distinguimos com precis\u00e3o as diferentes maneiras pelas quais uma solu\u00e7\u00e3o pode ser compreendida: como solu\u00e7\u00e3o expl\u00edcita ou impl\u00edcita, estendida ou maximal, e ressaltamos a import\u00e2ncia \u2014frequentemente subestimada\u2014 de declarar adequadamente seu dom\u00ednio. Essas distin\u00e7\u00f5es n\u00e3o s\u00e3o apenas formais: s\u00e3o operacionais. Ignor\u00e1-las pode nos levar, como vimos, a erros conceituais severos ao interpretar os resultados obtidos.<\/p>\n<p>Com este encerramento, j\u00e1 temos uma primeira ferramenta bem afiada. A compreens\u00e3o de uma EDO n\u00e3o pode se limitar a resolver uma f\u00f3rmula: requer senso cr\u00edtico, aten\u00e7\u00e3o aos detalhes e uma base conceitual s\u00f3lida que permita avan\u00e7ar sem perder o fio. Este \u00e9 apenas o come\u00e7o.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>O que \u00e9 uma Equa\u00e7\u00e3o Diferencial Ordin\u00e1ria (EDO)? Resumo:Nesta aula, exploram-se as Equa\u00e7\u00f5es Diferenciais Ordin\u00e1rias (EDO) de ordem k, come\u00e7ando com sua defini\u00e7\u00e3o e sua representa\u00e7\u00e3o de maneira normal e geral. Atrav\u00e9s de conceitos como a matriz Jacobiana e o Teorema da Fun\u00e7\u00e3o Impl\u00edcita, constroem-se as bases para compreender as solu\u00e7\u00f5es dessas equa\u00e7\u00f5es e as [&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":19,"footnotes":""},"categories":[1152,571,15],"tags":[],"class_list":["post-32846","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-equacoes-diferenciais-ordinarias","category-matematica-pt","category-sin-categoria-pt"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>O que \u00e9 uma Equa\u00e7\u00e3o Diferencial Ordin\u00e1ria (EDO)? - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Descubra o que \u00e9 uma Equa\u00e7\u00e3o Diferencial 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