Ejercicio: 2Eva2018TII_T2 EDO d2x/dt2 Kunge Kutta 2do Orden
\frac{\delta ^2 x}{\delta t^2} + 5t\frac{\delta x}{\delta t} +(t+7)\sin (\pi t) = 0 x'' + 5tx' +(t+7)\sin (\pi t) = 0 x'' = -5tx' -(t+7)\sin (\pi t) = 0si se usa z=x'
z' = -5tz -(t+7)\sin (\pi t) = 0se convierte en:
f(t,x,z) = x' = z g(t,x,z) = x'' = z' = -5tz -(t+7)sin (\pi t) = 0Tarea: Desarrollar 3 iteraciones en Papel.
Donde se aplica el algoritmo de Runge-Kutta d2y/d2x
t, x, z
[[ 0. 6. 1.5 ]
[ 0.2 6.3 0.92679462]
[ 0.4 6.38218195 -0.27187703]
[ 0.6 6.19792527 -1.17287944]
[ 0.8 5.88916155 -1.23638799]
[ 1. 5.6491005 -0.61819399]
[ 1.2 5.5872811 0.17288691]
[ 1.4 5.69750883 0.69945284]
[ 1.6 5.8992535 0.77223688]
[ 1.8 6.09372469 0.43437943]
[ 2. 6.20586248 -0.12630953]]
Instrucciones en Python
# 2Eva_IIT2018_T2 Kunge Kutta 2do Orden x''
import numpy as np
def rungekutta2_fg(f,g,x0,y0,z0,h,muestras):
tamano = muestras + 1
estimado = np.zeros(shape=(tamano,3),dtype=float)
# incluye el punto [x0,y0]
estimado[0] = [x0,y0,z0]
xi = x0
yi = y0
zi = z0
for i in range(1,tamano,1):
K1y = h * f(xi,yi,zi)
K1z = h * g(xi,yi,zi)
K2y = h * f(xi+h, yi + K1y, zi + K1z)
K2z = h * g(xi+h, yi + K1y, zi + K1z)
yi = yi + (K1y+K2y)/2
zi = zi + (K1z+K2z)/2
xi = xi + h
estimado[i] = [xi,yi,zi]
return(estimado)
# PROGRAMA
# INGRESO
f = lambda t,x,z: z
g = lambda t,x,z: -5*t*z-(t+7)*np.sin(np.pi*t)
t0 = 0
x0 = 6
z0 = 1.5
h = 0.2
muestras = 10
# PROCEDIMIENTO
tabla = rungekutta2_fg(f,g,t0,x0,z0,h,muestras)
# SALIDA
print(tabla)
# GRAFICA
import matplotlib.pyplot as plt
plt.plot(tabla[:,0],tabla[:,1])
plt.xlabel('t')
plt.ylabel('x(t)')
plt.show()