{"id":4611,"date":"2017-12-09T11:05:32","date_gmt":"2017-12-09T16:05:32","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=4611"},"modified":"2026-04-05T20:54:05","modified_gmt":"2026-04-06T01:54:05","slug":"s3eva2009tii_t2-edo-dy-dx-con-valor-inicial","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s3eva10\/s3eva2009tii_t2-edo-dy-dx-con-valor-inicial\/","title":{"rendered":"s3Eva2009TII_T2 EDO dy\/dx con Valor inicial, Runge-Kutta 4to orden"},"content":{"rendered":"\n

Ejercicio<\/strong><\/em>: 3Eva2009TII_T2 EDO dy\/dx con Valor inicial, Runge-Kutta 4to orden<\/a><\/p>\n\n\n\n

La ecuaci\u00f3n del problema:<\/p>\n\n\n (1-x^2)y' - xy = x (1-x^2) <\/span>\n\n\n\n

se despeja:<\/p>\n\n\n (1-x^2)y' = x (1-x^2) - xy <\/span>\n\n\n y' = x - \\frac{xy}{(1-x^2)} <\/span>\n\n\n\n

con valores iniciales x0<\/sub> = 0 , y0<\/sub> = 2, h = 0.1 en el intervalo [0, 0.5]<\/p>\n\n\n\n

Usando Runge-Kutta de 2do Orden<\/p>\n\n\n\n

Iteraci\u00f3n 1<\/h4>\n\n\n K_1 = h y'(0,2) <\/span>\n\n\n= (0.1)\\Big[0 - \\frac{0 (2)}{(1-(0^2))}\\Big] = 0 <\/span>\n\n\n K_2 = h y'(0+0.1, 2 + 0) <\/span>\n\n\n= (0.1)\\Big[0.1 - \\frac{0.1 (2+0)}{(1-(0.1^2))}\\Big] = 0.0302 <\/span>\n\n\ny_1= 2 + \\frac{0+0.0302}{2} = 2.0151<\/span>\n\n\nx_1 = x_0 + h = 0 + 0.1 = 0.1<\/span>\n\n\n\n

Iteraci\u00f3n 2<\/h4>\n\n\n K_1 = h y'(0.1,2.0151) <\/span>\n\n\n= (0.1)\\Big[0.1 - \\frac{0.1 (2.0151)}{(1-(0.1^2))}\\Big] = 0.0304 <\/span>\n\n\n K_2 = h y'(0.1+0.1, 2.0151 + 0.0304) <\/span>\n\n\n = (0.1)\\Big[0.2 - \\frac{0.2 (2.0151 + 0.0304)}{(1-(0.2^2))}\\Big] = 0.0626 <\/span>\n\n\ny_2 = 2.0151 + \\frac{0.0304+0.0626}{2} = 2.0616<\/span>\n\n\nx_2 = x_1 + h = 0.1 + 0.1 = 0.2<\/span>\n\n\n\n

Iteraci\u00f3n 3<\/h4>\n\n\n K_1 = h y'(0.2,2.0616) <\/span>\n\n\n= (0.1)\\Big[0.2 - \\frac{0.2 (2.0616)}{(1-(0.2^2))}\\Big] = 0.0629 <\/span>\n\n\n K_2 = h y'(0.2+0.1, 2.0616 + 0.0629) <\/span>\n\n\n = (0.1)\\Big[0.3 - \\frac{0.3 (2.0616 + 0.0629)}{(1-(0.3^2))}\\Big] = 0.1 <\/span>\n\n\ny_3 = 2.0151 + \\frac{0.0629+0.1}{2} = 2.1431<\/span>\n\n\nx_3 = x_2 + h = 0.2 + 0.1 = 0.3<\/span>\n\n\n\n

siguiendo el algoritmo se completa la tabla:<\/p>\n\n\n\n

 [xi,    yi,    K1,    K2    ]\n[[0.     2.     0.     0.    ]\n [0.1    2.0151 0.     0.0302]\n [0.2    2.0616 0.0304 0.0626]\n [0.3    2.1431 0.0629 0.1   ]\n [0.4    2.2668 0.1007 0.1468]\n [0.5    2.4463 0.1479 0.211 ]]<\/code><\/pre>\n\n\n\n
\"3eva2009tii_t2<\/figure>\n\n\n\n
Algoritmo en Python<\/h2>\n\n\n
\n# 3Eva_IIT2009_T2 Valor inicial Runge-Kutta 4to orden dy\/dx\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# INGRESO\nd1y = lambda x,y : x + x*y\/(1-x**2)\nx0 = 0\ny0 = 2\nh = 0.1\na = 0\nb = 1\/2\n\n# PROCEDIMIENTO\nmuestras = int((b -a)\/h)+1\ntabla = np.zeros(shape=(muestras,4),dtype=float)\n\ni = 0\nxi = x0\nyi = y0\ntabla[i,:] = [xi,yi,0,0]\n\ni = i+1\nwhile not(i>=muestras):\n    K1 = h* d1y(xi,yi)\n    K2 = h* d1y(xi+h,yi+K1)\n    yi = yi + (K1+K2)\/2\n    xi = xi +h\n    tabla[i,:] = [xi,yi,K1,K2]\n    i = i+1\n# vector para gr\u00e1fica\nxg = tabla[:,0]\nyg = tabla[:,1]\n\n# SALIDA\n# muestra 4 decimales\nnp.set_printoptions(precision=4)\nprint(' [xi, yi, K1, K2]')\nprint(tabla)\n\n# Gr\u00e1fica\nplt.plot(xg,yg)\nplt.xlabel('xi')\nplt.ylabel('yi')\nplt.grid()\nplt.show()\n<\/pre><\/div>\n\n\n
Tarea<\/strong><\/em>: Realizar con Runge-Kutta de 4to Orden<\/p>\n","protected":false},"excerpt":{"rendered":"
Ejercicio: 3Eva2009TII_T2 EDO dy\/dx con Valor inicial, Runge-Kutta 4to orden La ecuaci\u00f3n del problema: se despeja: con valores iniciales x0 = 0 , y0 = 2, h = 0.1 en el intervalo [0, 0.5] Usando Runge-Kutta de 2do Orden Iteraci\u00f3n 1 Iteraci\u00f3n 2 Iteraci\u00f3n 3 siguiendo el algoritmo se completa la tabla: Algoritmo 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":[50],"tags":[58,54],"class_list":["post-4611","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\/4611","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=4611"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4611\/revisions"}],"predecessor-version":[{"id":23912,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4611\/revisions\/23912"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=4611"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=4611"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=4611"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}