{"id":4596,"date":"2017-12-10T08:05:21","date_gmt":"2017-12-10T13:05:21","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=4596"},"modified":"2026-04-05T20:53:05","modified_gmt":"2026-04-06T01:53:05","slug":"s3eva2010ti_t2-edo-problema-con-valor-inicial-dy-dx","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s3eva10\/s3eva2010ti_t2-edo-problema-con-valor-inicial-dy-dx\/","title":{"rendered":"s3Eva2010TI_T2 EDO problema con valor inicial dy\/dx"},"content":{"rendered":"\n

Ejercicio<\/strong><\/em>: 3Eva2010TI_T2 EDO problema con valor inicial dy\/dx<\/a><\/p>\n\n\n\n

Ecuaci\u00f3n del problema:<\/p>\n\n\n (2y^2 + 4x^2)\\delta x -xy \\delta y =0 <\/span>\n\n\n\n

se despeja dy\/dx:<\/p>\n\n\n (2y^2 + 4x^2)\\delta x = xy \\delta y <\/span>\n\n\n \\frac{\\delta y}{\\delta x} =\\frac{2y^2 + 4x^2}{xy} <\/span>\n\n\n\n

con valores iniciales de x0<\/sub> = 1, y0<\/sub>=-2 , h=0.2 e intervalo [1,2]<\/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\\frac{\\delta y}{\\delta x}(1,-2)<\/span>\n\n\n =(0.2)\\frac{2(-2)^2 + 4(1)^2}{(1)(-2)}= -1.2 <\/span>\n\n\n K_2= h\\frac{\\delta y}{\\delta x}(1+0.2,-2+(-1.2))<\/span>\n\n\n =(0.2)\\frac{2(-2-1.2)^2 + 4(1+0.2)^2}{(1+0.2)(-2-1.2)}= -1.3667 <\/span>\n\n\n y_1 = -2 + \\frac{-1.2-1.3667}{2} = -3.2833<\/span>\n\n\n x_1 = x_0 + h = 1 + 0.2 = 1.2 <\/span>\n\n\n error = O(h^3) = O(0.2^3) = 0.008 <\/span>\n\n\n\n

Iteraci\u00f3n 2<\/h4>\n\n\n K_1= h\\frac{\\delta y}{\\delta x}(1.2,-3.2833)<\/span>\n\n\n =(0.2)\\frac{2(-3.2833)^2 + 4(1.2)^2}{(1.2)(-3.2833)}= -1.3868 <\/span>\n\n\n K_2= h\\frac{\\delta y}{\\delta x}(1.2+0.2,-3.2833+(-1.3868))<\/span>\n\n\n =(0.2)\\frac{2(-3.2833+(-1.3868))^2 + 4(1.2+0.2)^2}{(1.2+0.2)(-3.2833+(-1.3868))}= -1.5742 <\/span>\n\n\n y_2 = -3.2833 + \\frac{-1.3868-1.5742}{2} = -4.7638<\/span>\n\n\n x_2 = x_1 + h = 1.2 + 0.2 = 1.4 <\/span>\n\n\n\n

Iteraci\u00f3n 3<\/h4>\n\n\n K_1= h\\frac{\\delta y}{\\delta x}(1.4,-4.7638)<\/span>\n\n\n =(0.2)\\frac{2(-4.76383)^2 + 4(1.4)^2}{(1.4)(-4.7638)}= -1.5962 <\/span>\n\n\n K_2= h\\frac{\\delta y}{\\delta x}(1.4+0.2,-4.7638+(-1.5962))<\/span>\n\n\n =(0.2)\\frac{2(-4.7638+(-1.5962))^2 + 4(1.4+0.2)^2}{(1.4+0.2)(-4.7638+(-1.5962))}= -1.7913 <\/span>\n\n\n y_3 = -4.7638 + \\frac{-1.5962-1.7913}{2} = -6.4576<\/span>\n\n\n x_3 = x_2 + h = 1.4 + 0.2 = 1.6 <\/span>\n\n\n\n

con lo que usando el algoritmo se obtiene la tabla y gr\u00e1fica:<\/p>\n\n\n\n

 [xi,       yi,       K1,     K2     ]\n[[  1.      -2.       0.       0.    ]\n [  1.2     -3.2833  -1.2     -1.3667]\n [  1.4     -4.7638  -1.3868  -1.5742]\n [  1.6     -6.4576  -1.5962  -1.7913]\n [  1.8     -8.3698  -1.8126  -2.0119]\n [  2.     -10.5029  -2.032   -2.2342]]<\/code><\/pre>\n\n\n\n
\"3eva2010ti_t2<\/figure>\n\n\n\n
Algoritmo en Python<\/h2>\n\n\n
\n# 3Eva_IT2010_T2 EDO\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n# INGRESO\nd1y = lambda x,y : (2*(y**2)+4*(x**2))\/(x*y)\nx0 = 1\ny0 = -2\nh = 0.2\na = 1\nb = 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>: Realizar con Runge Kutta de 4to Orden<\/p>\n","protected":false},"excerpt":{"rendered":"
Ejercicio: 3Eva2010TI_T2 EDO problema con valor inicial dy\/dx Ecuaci\u00f3n del problema: se despeja dy\/dx: con valores iniciales de x0 = 1, y0=-2 , h=0.2 e intervalo [1,2] Usando Runge-Kutta de 2do Orden Iteraci\u00f3n 1 Iteraci\u00f3n 2 Iteraci\u00f3n 3 con lo que usando el algoritmo se obtiene la tabla y gr\u00e1fica: Algoritmo en Python Tarea: Realizar […]<\/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-4596","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\/4596","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=4596"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4596\/revisions"}],"predecessor-version":[{"id":23910,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4596\/revisions\/23910"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=4596"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=4596"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=4596"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}