Question

Question 22 1 pts Problem 22: Numerical solution of Ordinary differential equations Consider the following initial value problem GE:+15y = 1.C:y(0) -0.5 Carry out two-steps of the modified Euler (trapezoidal) method solution from the initial condition with a time step of At = 0.1. and the predicted solutions is y(0.2)-0.20 None of the above. y(0.2) - -0.75 y(0.2)-1.27 y(0.2)=0.25

Answer 1

Modified keelel Trapezoi dal ) melleod is aho Kuowon Reuge Katta- and order method 0 h=At=ool indement dy dt = [t- 15y = f(tig) 7(0) = 0.5 so, to=o. Yo = or for modified enter (traperoidal) method Yitia yot 0.5h [titi Alti, Yi)t ftithe, Yit hati, 5, 43) Y = Tot 0.5th [ f( to, yo) + Altoth, Yet ht(to yo)] flto, 4o) = t-154 ) 3,- 054 05 404(+5+ 401, 034 01 x 14,5)}] 75+ F ( ool, 0.5+ 0.1 x6-745) y, = 0.5+ 0.05 C - -(-705f f (001,-0.25 0 - 15X0.5 = -7.5 Yo= 0.5 to=o h=Oul f(0-1,-0.25) = s) Ool- 15X-0.25 3.85 1 = 0.5+ 0.05 ( -705+ 3.85] 0-3175 - Y = 0.3175 ti= to th= 0 +0 1 = 0.1 ftiy)= t-ity flt gy)= Ool - 15X0.3175 Ht,,y) = 0.1- 4.7625 =~4.6625 Ht,y)= -4.6625 9, = 0.3175 tia Oul

2 2 = 0.3175+ 0.1 I 2 = 4, + & [ +(+1,91) + f(tith, Yith flti, Yi. yo)] -4.6625+ f ( 0.17 0.1, 0.3175+ 0.1 *_4.4625)] =0.3175+ ovos -4.6625+ f(0.2, -0.14875) as fl0.2,-0.14875) = 0.2- 15x(-0.14875) O.. [- 2013125 Y2 = 0.3175+ 0.05 y ( -4.66 25+ P243125 0:3175 Oo11156 Ya = 0.2059 So at t=012, Y = 0.2059 y(0.2) = 0.20 option ①