Question

Consider the initial value problem below to answer to following. a) Find the approximations to y(0.3) and y(0.6) using Euler's method with time steps of At 0.3, 0.15, 0.075, and 0.0375 b) Using the exact solution given, compute the errors in the Euler approximations at t 0.3 and t 0.6. c) Which time step results in the more accurate approximation? Explain your observations. d) In general, how does halving the time step affect the error at t 0.3 and t 0.6? y')-2y. y(0)-1. y)e2 a) Complete the table below. At Approximation y(0.3) Approximation to y(0.6) to 0.3 0.15 0.075 0.0375 (Round to five decimal places as needed) b) Complete the table below Errors for y(0.3) Errors for y(0.6) At 0.3 0.15 0.075 0.0375 (Round to five decimal places as needed) c) Which time step results in the more accurate approximation? OA. Time step At 0.3 results in the more accurate approximation because larger time steps generally produce smaller numbers O B. Time step At 0.3 results in the more accurate approximation because larger time steps generally produce more accurate results O C. Time step At 0.0375 results in the more accurate approximation because smaller time steps generally produce more accurate results O D. Time step At 0.0375 results in the more accurate approximation because smaller time steps generally produce smaller numbers. d) How did halving the time step affect the error at t 0 3 and t 0.6? O A. Halving the time steps results in reducing the eror by approximately one fourth O B. Halving the time steps results in approximately halving the error O C. Halving the time steps results in approximately doubling the error O D. Halving the time steps results in reducing the error by approximately one tenth Click to select your answer(S) Save for Later

Answer 1

"Euler's method: - y_n + 1 = y_n + hf (x_n, y_n) rightarrow Iterative formula where dy/dx = f (x, y): y (x_0) = y_0 is the initial guess h = step size Given y' (t) = f' (t) = dy/dt = -2y = f (t, y): y (0) = 1 Implies downarrow: t_0 = 0: y_0 = 1 rightarrow Delta t = h = step size = 0.3 y (0.3) = y (0 + 0.3) = y (x_0 + Delta t) = y (x_0 + h) = y (x_1) = y_1 Implies y_1 = y_0 + h f (t_0, y_0) = 1 + 0.3 f (0, 1) y_1 = 1 + 0.3 (-2) = 1 - 0.6 = 0.4 Therefore y_1 = 0.4 Exact solution y = e^-2t Implies y (0.3) = e^-2(0.3) = e^-0.6 = 0.5488 Therefore Error = Exact value - approximate value = 0.5488 - 0.4 Therefore Error = 0.1488 rightarrow (1) Absolute Error =