{"id":52,"date":"2017-11-06T12:42:07","date_gmt":"2017-11-06T11:42:07","guid":{"rendered":"http:\/\/clases.jesussoto.es\/?p=52"},"modified":"2017-11-06T12:42:07","modified_gmt":"2017-11-06T11:42:07","slug":"efm-ed-lineales-de-primer-orden","status":"publish","type":"post","link":"https:\/\/curso17.jesussoto.es\/?p=52","title":{"rendered":"EFM: ED lineales de primer orden"},"content":{"rendered":"<p>p>El pasado d\u00eda dec\u00edamos que la ecuaci\u00f3n diferencial $P( x, y)\\, dx + Q(x, y)\\, dy = 0$ era exacta si $\\frac{\\partial P}{\\partial y}=\\frac{\\partial Q}{\\partial x}$. Aprendimos a resolver este tipo de ecuaciones. Sin embargo podemos toparnos con ecuaciones que no lo cumplan, pero que al multiplicarles determinada funci\u00f3n, $\\mu$, verifique $$\\frac{\\partial \\mu P}{\\partial y}=\\frac{\\partial \\mu Q}{\\partial x}.$$<\/p>\n<p>A esta funci\u00f3n $\\mu$, la denominamos factor integrante. A veces, el uso de factores integrantes nos ayudan a simplificar una ecuaci\u00f3n diferencial (ED). En general la transformamos en una ED exacta o en una ED de variables separadas.<\/p>\n<p>Esta t\u00e9cnica nos permite resolver la ED<\/p>\n<p>$$\\begin{cases} y\u2019+P(x)y = Q(x)\\\\ y(x_0) = y_0 \\end{cases}$$<\/p>\n<p>Donde $P(x)$ y $Q(x)$ son funciones continuas en un intervalo $[a,b] \\subseteq \\mathbb{R}$.<\/p>\n<p>Si utilizamos el factor integrante<\/p>\n<p>$$e^{\\int_{x_0}^x P(x) dx }, $$<\/p>\n<p>la soluci\u00f3n de esta ecuaci\u00f3n viene dada por:<\/p>\n<p>$$y(x) =e^{ \u2013 \\int_{x_0}^x P(x) dx } \\left[ y_0 + \\int_{x_0}^x Q(x) e^{ \\int P(x) dx } dx \\right]$$<\/p>\n<p><!--  --><\/p>\n<table id=\"yzpi\" width=\"100%\" border=\"0\" cellspacing=\"0\" cellpadding=\"3\" bgcolor=\"#999999\">\n<tbody>\n<tr>\n<td width=\"100%\"><strong>Ejercicio:<\/strong>Resolver $y\u2019\\,\\cos(x)+y\\sin(x)-1=0$<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n","protected":false},"excerpt":{"rendered":"<p>p>El pasado d\u00eda dec\u00edamos que la ecuaci\u00f3n diferencial $P( x, y)\\, dx + Q(x, y)\\, dy = 0$ era exacta si $\\frac{\\partial P}{\\partial y}=\\frac{\\partial Q}{\\partial x}$. Aprendimos a resolver este tipo de ecuaciones. Sin embargo podemos toparnos con ecuaciones que no lo cumplan, pero que al multiplicarles determinada funci\u00f3n, $\\mu$, verifique $$\\frac{\\partial \\mu P}{\\partial y}=\\frac{\\partial&hellip; <a class=\"more-link\" href=\"https:\/\/curso17.jesussoto.es\/?p=52\">Seguir leyendo <span class=\"screen-reader-text\">EFM: ED lineales de primer orden<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[3],"tags":[],"_links":{"self":[{"href":"https:\/\/curso17.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/52"}],"collection":[{"href":"https:\/\/curso17.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/curso17.jesussoto.es\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/curso17.jesussoto.es\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/curso17.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=52"}],"version-history":[{"count":1,"href":"https:\/\/curso17.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/52\/revisions"}],"predecessor-version":[{"id":53,"href":"https:\/\/curso17.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/52\/revisions\/53"}],"wp:attachment":[{"href":"https:\/\/curso17.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=52"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/curso17.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=52"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/curso17.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=52"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}