{"id":4551,"date":"2017-12-07T11:10:27","date_gmt":"2017-12-07T16:10:27","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=4551"},"modified":"2026-04-05T20:55:09","modified_gmt":"2026-04-06T01:55:09","slug":"s3eva2007tii_t3-edo-dy-dt-taylor-orden-2","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s3eva10\/s3eva2007tii_t3-edo-dy-dt-taylor-orden-2\/","title":{"rendered":"s3Eva2007TII_T3 EDO dy\/dt Taylor orden 2"},"content":{"rendered":"\n

Ejercicio<\/strong><\/em>: 3Eva2007TII_T3 EDO dy\/dt Taylor orden 2<\/a><\/p>\n\n\n\n

La ecuaci\u00f3n del problema es:<\/p>\n\n\n y'= 1 +\\frac{y}{t} + \\Big(\\frac{y}{t}\\Big) ^2 <\/span>\n\n\n 1\\leq t\\leq 2<\/span>\n\n\n y(1)=0, h=0.2<\/span>\n\n\n\n

Tomando como referencia Taylor de 3 t\u00e9rminos m\u00e1s el t\u00e9rmino de error O(h3<\/sup>)<\/p>\n\n\n y_{i+1} = y_i +\\frac{h}{1!}y'_i + \\frac{h^2}{2!}y''_i + \\frac{h^3}{3!}y'''_i<\/span>\n\n\n\n

Se usa hasta el tercer t\u00e9rmino para el algoritmo.<\/p>\n\n\n y_{i+1} = y_i +\\frac{h}{1!}y'_i + \\frac{h^2}{2!}y''_i<\/span>\n\n\n\n

Se determina que se requiere la segunda derivada para completar la aproximaci\u00f3n. A partir de la ecuaci\u00f3n del problema se aplica en cada t\u00e9rmino:<\/p>\n\n\n \\Big(\\frac{u}{v}\\Big)' = \\frac{u'v-uv' }{v^2}<\/span>\n\n\n y'= 1 +\\frac{y}{t} + \\frac{y^2}{t^2} <\/span>\n\n\n y''= 0+\\frac{y't-y}{t^2} + \\frac{2yy't^2-2y^2t}{t^4} <\/span>\n\n\n y''= \\frac{y't-y}{t^2} + \\frac{2y\\Big(1 +\\frac{y}{t} + \\frac{y^2}{t^2}\\Big) t^2-2y^2t}{t^4} <\/span>\n\n\n\n

Con lo que se realizan las iteraciones para llenar la tabla<\/p>\n\n\n\n

iteraci\u00f3n 1<\/strong><\/em><\/p>\n\n\n\n

t0<\/sub> = 1<\/p>\n\n\n\n

y0<\/sub>= 0<\/p>\n\n\n y'= 1 +\\frac{0}{1} + \\Big(\\frac{0}{1}\\Big)^2 = 1<\/span>\n\n\n y''= \\frac{(1)(1)-0}{(1)^2} + \\frac{2(0)(1)(1)^2-2(0)^2(1)}{(1)^4} = 1<\/span>\n\n\n y_{1} = 0 +\\frac{0.2}{1!}(1) + \\frac{0.2^2}{2!}(1) = 0.22<\/span>\n\n\n\n

t1<\/sub> = t0<\/sub> + h = 1 + 0.2 = 1.2<\/p>\n\n\n\n

iteraci\u00f3n 2<\/strong><\/em><\/p>\n\n\n\n

t1<\/sub> = 1.2<\/p>\n\n\n\n

y1<\/sub>= 0.22<\/p>\n\n\n y'= 1 +\\frac{0.22}{1.2} + \\Big(\\frac{0.22}{1.2}\\Big) ^2 = 1.21694444 <\/span>\n\n\n y''= \\frac{y'(1.2)-0.22}{1.2^2} + \\frac{2(0.22)y'(1.2)^2-2(0.22)^2(1.2)}{(1.2)^4} = 1.17716821 <\/span>\n\n\n y_{2} = 0.22 +\\frac{0.2}{1!}(1.21694444) + \\frac{0.2^2}{2!}(1.17716821) = 0.48693225<\/span>\n\n\n\n

t2<\/sub> = t1<\/sub> + h = 1.2 + 0.2 = 1.4<\/p>\n\n\n\n

iteraci\u00f3n 3<\/strong><\/em><\/p>\n\n\n\n

t2<\/sub> = 1.4<\/p>\n\n\n\n

y2<\/sub>= 0.48693225<\/p>\n\n\n y'= 1 +\\frac{0.48693225}{1.4} + \\Big(\\frac{0.48693225}{1.4}\\Big) ^2 = 1.46877968<\/span>\n\n\n y''= \\frac{(1.46877968)(1.4)-0.48693225}{1.4^2} + <\/span>\n\n\n\\frac{2(0.48693225)(1.46877968)(1.4)^2-2(0.48693225)^2(1.4)}{1.4^4} = 1.35766995 <\/span>\n\n\n y_{3} = 0.48693225 +\\frac{0.2}{1!}(1.46877968) + \\frac{0.2^2}{2!}(1.35766995) = 0.80784159<\/span>\n\n\n\n

t3<\/sub> = t2<\/sub> + h = 1.4 + 0.2 = 1.6<\/p>\n\n\n\n

Con lo que se puede realizar el algoritmo, obteniendo la siguiente tabla:<\/p>\n\n\n\n

estimado\n[xi,        yi,        d1yi,      d2yi]\n[[1. 0. 1. 1. ] \n [1.2 0.22 1.21694444 1.17716821] \n [1.4 0.48693225 1.46877968 1.35766995] \n [1.6 0.80784159 1.759826 1.57634424] \n [1.8 1.19133367 2.09990017 1.8564435 ] \n [2. 1.64844258 0. 0. ]]<\/code><\/pre>\n\n\n\n
\"3Eva_IIT2007_T3<\/figure>\n","protected":false},"excerpt":{"rendered":"
Ejercicio: 3Eva2007TII_T3 EDO dy\/dt Taylor orden 2 La ecuaci\u00f3n del problema es: Tomando como referencia Taylor de 3 t\u00e9rminos m\u00e1s el t\u00e9rmino de error O(h3) Se usa hasta el tercer t\u00e9rmino para el algoritmo. Se determina que se requiere la segunda derivada para completar la aproximaci\u00f3n. A partir de la ecuaci\u00f3n del problema se aplica […]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn-ejemplo","format":"standard","meta":{"footnotes":""},"categories":[50],"tags":[58,54],"class_list":["post-4551","post","type-post","status-publish","format-standard","hentry","category-mn-s3eva10","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4551","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=4551"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4551\/revisions"}],"predecessor-version":[{"id":23915,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4551\/revisions\/23915"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=4551"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=4551"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=4551"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}