{"id":3192,"date":"2019-01-31T10:14:06","date_gmt":"2019-01-31T15:14:06","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=3192"},"modified":"2026-04-05T20:12:43","modified_gmt":"2026-04-06T01:12:43","slug":"s2eva2018tii_t2-edo-d2xdt2-kunge-kutta-2orden","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva20\/s2eva2018tii_t2-edo-d2xdt2-kunge-kutta-2orden\/","title":{"rendered":"s2Eva2018TII_T2 EDO d2x\/dt2 Kunge Kutta 2do Orden"},"content":{"rendered":"\n<p><em><strong>Ejercicio<\/strong><\/em>: <a href=\"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-2eva20\/2eva2018tii_t2-edo-d2xdt2-kunge-kutta-2orden\/\" data-type=\"post\" data-id=\"3144\">2Eva2018TII_T2 EDO d2x\/dt2 Kunge Kutta 2do Orden<\/a><\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> \\frac{\\delta ^2 x}{\\delta t^2} + 5t\\frac{\\delta x}{\\delta t} +(t+7)\\sin (\\pi t) = 0 <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> x'' + 5tx' +(t+7)\\sin (\\pi t) = 0 <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> x'' = -5tx' -(t+7)\\sin (\\pi t) = 0 <\/span>\n\n\n\n<p>si se usa z=x'<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> z' = -5tz -(t+7)\\sin (\\pi t) = 0 <\/span>\n\n\n\n<p>se convierte en:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\">f(t,x,z) = x' = z <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\">g(t,x,z) = x'' = z' = -5tz -(t+7)sin (\\pi t) = 0 <\/span>\n\n\n\n<p><strong>Tarea<\/strong>: Desarrollar 3 iteraciones en Papel.<\/p>\n\n\n\n<p>Donde se aplica el algoritmo de <a href=\"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-unidades\/mn-u06\/runge-kutta-d2y-dx2\/\" data-type=\"post\" data-id=\"1989\">Runge-Kutta d2y\/d2x<\/a><\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>   t,              x,              z\n&#091;&#091; 0.          6.          1.5       ]\n &#091; 0.2         6.3         0.92679462]\n &#091; 0.4         6.38218195 -0.27187703]\n &#091; 0.6         6.19792527 -1.17287944]\n &#091; 0.8         5.88916155 -1.23638799]\n &#091; 1.          5.6491005  -0.61819399]\n &#091; 1.2         5.5872811   0.17288691]\n &#091; 1.4         5.69750883  0.69945284]\n &#091; 1.6         5.8992535   0.77223688]\n &#091; 1.8         6.09372469  0.43437943]\n &#091; 2.          6.20586248 -0.12630953]]<\/code><\/pre>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"638\" height=\"479\" src=\"http:\/\/blog.espol.edu.ec\/algoritmos101\/files\/2019\/01\/2EIIT2018T2_Kunge-Kutta_2do-Orden_x.png\" alt=\"2eiit2018t2 Runge-Kutta 2do orden x\" class=\"wp-image-18514\" \/><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Instrucciones en Python<\/h2>\n\n\n<div class=\"wp-block-syntaxhighlighter-code \"><pre class=\"brush: python; title: ; notranslate\" title=\"\">\n# 2Eva_IIT2018_T2 Kunge Kutta 2do Orden x''\nimport numpy as np\n\ndef rungekutta2_fg(f,g,x0,y0,z0,h,muestras):\n    tamano = muestras + 1\n    estimado = np.zeros(shape=(tamano,3),dtype=float)\n    # incluye el punto &#x5B;x0,y0]\n    estimado&#x5B;0] = &#x5B;x0,y0,z0]\n    xi = x0\n    yi = y0\n    zi = z0\n    for i in range(1,tamano,1):\n        K1y = h * f(xi,yi,zi)\n        K1z = h * g(xi,yi,zi)\n        \n        K2y = h * f(xi+h, yi + K1y, zi + K1z)\n        K2z = h * g(xi+h, yi + K1y, zi + K1z)\n\n        yi = yi + (K1y+K2y)\/2\n        zi = zi + (K1z+K2z)\/2\n        xi = xi + h\n        \n        estimado&#x5B;i] = &#x5B;xi,yi,zi]\n    return(estimado)\n\n# PROGRAMA\n# INGRESO\nf = lambda t,x,z: z\ng = lambda t,x,z: -5*t*z-(t+7)*np.sin(np.pi*t)\nt0 = 0\nx0 = 6\nz0 = 1.5\nh = 0.2\nmuestras = 10\n\n# PROCEDIMIENTO\ntabla = rungekutta2_fg(f,g,t0,x0,z0,h,muestras)\n\n# SALIDA\nprint(tabla)\n# GRAFICA\nimport matplotlib.pyplot as plt\nplt.plot(tabla&#x5B;:,0],tabla&#x5B;:,1])\nplt.xlabel('t')\nplt.ylabel('x(t)')\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"<p>Ejercicio: 2Eva2018TII_T2 EDO d2x\/dt2 Kunge Kutta 2do Orden si se usa z=x' se convierte en: Tarea: Desarrollar 3 iteraciones en Papel. Donde se aplica el algoritmo de Runge-Kutta d2y\/d2x Instrucciones en Python<\/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":[48],"tags":[58,54],"class_list":["post-3192","post","type-post","status-publish","format-standard","hentry","category-mn-s2eva20","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/3192","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=3192"}],"version-history":[{"count":4,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/3192\/revisions"}],"predecessor-version":[{"id":23859,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/3192\/revisions\/23859"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=3192"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=3192"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=3192"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}