Write Matlab function [T , Y] = rk4(f,tspan,y0,n)

that solves dy/dt = f(t,y) with initial condition y(t = a) = y0

on the interval [a,b] using n steps of the classical 4th order Runge-Kutta method. Inputs: f -- a function f(t,y) that returns a column vector of the same length as y tspan -- a vector [a,b] with the initial and end values of t y0 -- a column vector of the initial values n -- number of steps to use

Outputs: T -- a n+1 column vector containing the times Y -- a (n+1) by d matrix, where d is the length of y; Y(t,i) gives the i-th component of y at timestep t For example, [T , Y] = rk4(@(t, y) -y, [0 1], 1, 2) should return T = [0 0.5 1] Y = [1 0.60677 0.36817]

As another example, [T , Y] = rk4(@(t, y) [y(2); -y(1)], [0 1], [1; 0], 2) should return T = [0 0.5 1] Y = [1 0 0.87760 -0.47917 0.54059 -0.84104]