{"id":4756,"date":"2020-01-29T09:44:06","date_gmt":"2020-01-29T14:44:06","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=4756"},"modified":"2026-04-05T20:10:50","modified_gmt":"2026-04-06T01:10:50","slug":"s2eva2019tii_t2-edo-problema-de-valor-inicial","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva20\/s2eva2019tii_t2-edo-problema-de-valor-inicial\/","title":{"rendered":"s2Eva2019TII_T2 EDO problema de valor inicial"},"content":{"rendered":"\n

Ejercicio<\/strong>: 2Eva2019TII_T2 EDO problema de valor inicial<\/a><\/p>\n\n\n\n

la ecuaci\u00f3n del problema planteado es:<\/p>\n\n\n y'(t) = f(t,y) = \\frac{y}{2t^3} <\/span>\n\n\n\n

0 \u2264 t \u2264 1
y(0.5) = 1.5<\/p>\n\n\n\n

\n\n\n\n

literal a<\/h2>\n\n\n\n

La soluci\u00f3n empieza usando la Serie de Taylor por ejemplo para tres t\u00e9rminos:<\/p>\n\n\n y_{i+1} = y_{i} + h y'_i + \\frac{h^2}{2!} y''_i <\/span>\n\n\n x_{i+1} = x_{i} + h<\/span>\n\n\n E = \\frac{h^3}{3!} y'''(z) = O(h^3)<\/span>\n\n\n\n

Se observa que se tiene el valor inicial y la primera derivada, si usamos tres t\u00e9rminos se puede usar la segunda derivada.<\/p>\n\n\n y'(t) = f(t,y) = \\frac{y}{2t^3} <\/span>\n\n\n y''(t) = f'(t,y) = \\frac{y'}{2t^3} + y \\Big(\\frac{-3}{2t^4}\\Big)<\/span>\n\n\n y''(t) = \\frac{1}{2t^3}\\Big( \\frac{y}{2t^3} \\Big) - \\frac{3y}{2t^4}<\/span>\n\n\n y''(t) = \\frac{y}{4t^6} -\\frac{3y}{2t^4}<\/span>\n\n\n\n

por lo que la ecuaci\u00f3n de Taylor a usar queda de la siguiente forma:<\/p>\n\n\n y_{i+1} = y_{i} + h y'_i + \\frac{h^2}{2!} y''_i <\/span>\n\n\n y_{i+1} = y_{i} + h\\frac{y}{2t^3}+\\frac{h^2}{2} \\Big( \\frac{y}{4t^6} -\\frac{3y}{2t^4}\\Big) <\/span>\n\n\n\n

que es la ecuacion que se usar\u00e1 con un error de O(h3<\/sup>)<\/p>\n\n\n\n

Reemplazando los valores en la f\u00f3rmula se obtiene la siguiente tabla:<\/p>\n\n\n\n

estimado\n [ti,      yi,      d1yi,    d2yi]\n[[  0.5    1.5      6.      -12.        ]\n [  0.6    2.04     4.7222  -12.68004115]\n [  0.7    2.4488   3.5697  -10.09510219]\n [  0.8    2.7553   2.6907  -7.46259891]\n [  0.9    2.9870   2.0487  -5.42399041]\n [  1.     3.1648   0.       0.        ]]<\/code><\/pre>\n\n\n\n
cuya gr\u00e1fica es:<\/p>\n\n\n\n
\"2e2019tii_t2<\/figure>\n\n\n\n
\n\n\n\n
literal b<\/h2>\n\n\n\n
Para desarrollar Runge-Kutta de 2do orden se dispone de los siguientes datos:<\/p>\n\n\n y'(t) = f(t,y) = \\frac{y}{2t^3} <\/span>\n\n\n\n
t0<\/sub> = 0.5, y0<\/sub> = 1.5, h = 0.1<\/p>\n\n\n\n
pasos del algoritmo,<\/p>\n\n\n K_1 = h * y'(t_i)<\/span>\n\n\n K_2 = h * y'(t_i+h, y_i + K_1)<\/span>\n\n\n y_{i+1} = y_i + \\frac{K_1+K_2}{2}<\/span>\n\n\n t_i = t_i + h<\/p>\n<p><\/span>\n\n\n\n
iteraci\u00f3n 1: i = 0<\/p>\n\n\n K_1 = 0.1 * y'(0.5) = 0.6 <\/span>\n\n\n K_2 = 0.1 * y'(0.5+0.1, 1.5 + 0.6) = 0.4861 <\/span>\n\n\n y_{1} = 1.5 + \\frac{0.6+0.4861}{2} = 2.0430 <\/span>\n\n\n t_1 = 0.5 + 0.1 = 0.6<\/p>\n<p><\/span>\n\n\n\n
iteraci\u00f3n 2: i = 1<\/p>\n\n\n K_1 = 0.1 * y'(0.6) = 0.4729 <\/span>\n\n\n K_2 = 0.1 * y'(0.6+0.1, 2.0430 + 0.4729) = 0.3667 <\/span>\n\n\n y_{1} = 2.0430 + \\frac{0.4729+0.3667}{2} = 2.4629 <\/span>\n\n\n t_1 = 0.6 + 0.1 = 0.7<\/p>\n<p><\/span>\n\n\n\n
iteraci\u00f3n 3: i = 2<\/p>\n\n\n K_1 = 0.1 * y'(0.7) = 0.3590 <\/span>\n\n\n K_2 = 0.1 * y'(0.7+0.1, 2.4629 + 0.3590) = 0.3667 <\/span>\n\n\n y_{1} = 2.4629 + \\frac{0.3590+0.3667}{2} = 2.7802 <\/span>\n\n\n t_1 = 0.7 + 0.1 = 0.8<\/p>\n<p><\/span>\n\n\n\n
obteniendo la siguiente tabla:<\/p>\n\n\n\n
estimado\n [ti,   yi,     K1,     K2]\n[[0.5   1.5     0.      0.    ]\n [0.6   2.0430  0.6     0.4861]\n [0.7   2.4629  0.4729  0.3667]\n [0.8   2.7802  0.3590  0.2755]\n [0.9   3.0206  0.2715  0.2093]\n [1.    3.2048  0.2071  0.1613]]\nDiferencias entre Taylor y Runge-Kutta2\n[ 0.   -0.0030 -0.0140 -0.0248 -0.0335 -0.0400]<\/code><\/pre>\n\n\n\n
\n\n\n\n
Algoritmo en Python<\/h2>\n\n\n
\n# EDO. M\u00e9todo de Taylor 3 t\u00e9rminos\n# Runge-Kutta de 2 Orden\n# 2Eva_IIT2019_T2 EDO, problema de valor inicial\nimport numpy as np\n\n# Funciones desarrolladas en clase\ndef edo_taylor3t(d1y,d2y,x0,y0,h,muestras):\n    tamano = muestras + 1\n    estimado = np.zeros(shape=(tamano,4),dtype=float)\n    # incluye el punto [x0,y0]\n    estimado[0] = [x0,y0,0,0]\n    x = x0\n    y = y0\n    for i in range(1,tamano,1):\n        estimado[i-1,2:]= [d1y(x,y),d2y(x,y)]\n        y = y + h*d1y(x,y) + ((h**2)\/2)*d2y(x,y)\n        x = x+h\n        estimado[i,0:2] = [x,y]\n    return(estimado)\ndef rungekutta2(d1y,x0,y0,h,muestras):\n    tamano = muestras + 1\n    estimado = np.zeros(shape=(tamano,4),dtype=float)\n    # incluye el punto [x0,y0]\n    estimado[0] = [x0,y0,0,0]\n    xi = x0\n    yi = y0\n    for i in range(1,tamano,1):\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        estimado[i] = [xi,yi,K1,K2]\n    return(estimado)\n\n# PROGRAMA PRUEBA\n# INGRESO\n# d1y = y', d2y = y''\nd1y = lambda t,y: y\/(2*t**3)\nd2y = lambda t,y: y\/(4*t**6)-(3\/2)*y\/(t**4)\nt0 = 0.5\ny0 = 1.5\nh = 0.1\nmuestras = 5\n\n# PROCEDIMIENTO\n# Edo con Taylor\npuntos = edo_taylor3t(d1y,d2y,t0,y0,h,muestras)\nti = puntos[:,0]\nyi = puntos[:,1]\n\n# Runge-Kutta\npuntosRK2 = rungekutta2(d1y,t0,y0,h,muestras)\n# ti = puntosRK2[:,0] # lo mismo del anterior\nyiRK2 = puntosRK2[:,1]\n\n# diferencias\ndiferencia = yi-yiRK2\n\n# SALIDA\nprint('estimado[ti, yi, d1yi, d2yi]')\nprint(puntos)\n\nprint('estimado[ti, yi, K1, K2]')\nprint(puntosRK2)\n\nprint('Diferencias entre Taylor y Runge-Kutta2')\nprint(diferencia)\n\n# Gr\u00e1fica\nimport matplotlib.pyplot as plt\nplt.plot(ti[0],yi[0],'o', color='r',\n         label ='[t0,y0]')\nplt.plot(ti[1:],yi[1:],'o', color='g',\n         label ='y con Taylor 3 t\u00e9rminos')\nplt.plot(ti[1:],yiRK2[1:],'o', color='m',\n         label ='y Runge-Kutta 2 Orden')\nplt.title('EDO: Soluci\u00f3n num\u00e9rica')\nplt.xlabel('t')\nplt.ylabel('y')\nplt.legend()\nplt.grid()\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2019TII_T2 EDO problema de valor inicial la ecuaci\u00f3n del problema planteado es: 0 \u2264 t \u2264 1y(0.5) = 1.5 literal a La soluci\u00f3n empieza usando la Serie de Taylor por ejemplo para tres t\u00e9rminos: Se observa que se tiene el valor inicial y la primera derivada, si usamos tres t\u00e9rminos se puede usar la […]<\/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-4756","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\/4756","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=4756"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4756\/revisions"}],"predecessor-version":[{"id":23853,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4756\/revisions\/23853"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=4756"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=4756"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=4756"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}