Please answer and include matlab code

(By Hand) Suppose that an ordinary differential equation is Solved numerically on an interval [1:1, b] and that the local truncation error is chp. Show that if all truncation errors have the same sign (the worst possible case), then the total truncation error is (b a)chp_1, where h = [b a]. (By Hand) Consider the following ODE: at = u2 251mm) + [cos(2t])2, t 6 [0,1] MO) 2 1 Verify that n(t] = cos(2t] satises both the ODE and the initial condition. (Matlab Problem) Write a Matlab script to compute an Euler approximation to the above ODE using n = 10, 20, 40, 80 time steps (h = 0.1, 0.05, 0.025, 0.0125). Tabulate your results by showing four columns: the step size h, the value of the approxirnation at t = 1,. the error, and the observed order of convergence p for each value of h. Recall that the order of convergence p is computed as 2 W 103(hk/hk1) where err(hk) is the error at t = 1 corresponding to a step size of he. Do you see the expected convergence? If not, try to explain your results. p