Question

The outputs are very close to the correct values so I assume I just typed an equation incorrectly, but I have not been able to find the error.

Courses LMS integration Documentation Write a function to implement the 4th order Runge Kutta method function [ty] - r4(f, range, ich) where is an anonymous function that defines the differential equation range is a vector of length 2 that sets a time range for a range for the independent art the initial condition, and his a scalar step size i c le that Function Rose MATLAB Documentation function (ty) = 4(f.range, ic,h) 2 Xinput f is the RHS of the ODE Xinput range is a vector initial time, final time . Xinput it is an initial condition $ input his a step size zoutput t is an array containing the times that correspond to the approximations in y Zoutput y is an array containing approximations to the solution of the IVP sset up the time array 1 t (range (1) ihirange(2) 11 spreallocate for y array 12y -tros (length(t),1); 13 (1)ic; in for full length(t)-1 kif(y(),t()); yl-y(1) kih/2; 2-(y1.t()); y2-y(1) 2 /2 f(y2.t(1));

Answer 1

import math def f(x, y): b=X+y return (math.sqrt(b)) def rk4(f, rangel, ic, h): n = (int((x - range1[0][0])/h)) y = ic range1[0][1]=y for i in range(1, n+1): k1 = h * fly, range1[i-1][0]) k2 = h * f(y + 0.5 * k1,range1[i-1][0] + 0.5 * h) k3 = h * f(y + 0.5 * k2,range1[i-1][0] + 0.5 * h) k4 = h * f(y + k3, range1[i-1][0] + h) y = y + (1.0 / 6.0)*(K1 + 2 * k2 + 2 * k3 + k4) range1[i][1]=y return range1 X0 = 1 ic = 3 x = 2 h = 0.247 rows=0 cols=0 n=int((x-x0)/h) rows=n cols=2 print(n) range1 = [[@ for i in range(cols)] for j in range (rows+1)] range1[0][0]=xel i=1 while i<(int(x-x0)/h): range1[i][0]=range1[i-1][O]+h i=i+1 rk4(f, range1, ic, h)

just change the value of h to 0.247