Question

5. Consider the system of differential equations yi = y1 + 2y2, y = -41/2 + y2 with initial conditions yi(0) = 1, y2(0= 0. This has exact solution yı(t) = exp(t) cos(t), yz(t) = - exp(t) sin(t)/2. (a) Apply Euler's method with h=1/4 and find the global truncation error by comparing with the exact solution over the interval [0, 1]. (b) Apply the RK4 method with h=1 and find the global truncation error by comparing with the exact solution over the interval [0, 1]. Compare with the performance of Euler's method.

Answer 1

Enact nowtion Eulez and RA Y'z f(J., 92) = 9,+242 . 14, (O)=1 ₂ = 9 (9, 92) = -4/2 + 92 | 92 (O)=0 energia et)eup (4) Con (4), Y, (t) = - eup (t) si (4) Euler Method - Yo (tth) z Y, (t) & hof (9, (t) h 20.25 In (t thj - Y₂ (t) thg (9, (+) , Ya(t)) 4(0.25) = Y, COW+ 0.25(1 +2x07 21.25 y 0-25) 2 426) + 0.2042 +o) -.-0.125 Y, (0.5) > 4,1025)+ 0.2511125+ 24-0.120)=1.5- 92 (0.5) 2 92 (0.25) +0.251-1,25+(-0.1251) >-0.3125 Y, (0.75) - 9,60.5) +0.25/1572(-0.31139 26).71875 yg (0.75), Y, (0.5) +0.254-43 70.3125)) 2-0.578125 4,(!) z 9, (075) +0.25(1.71875-20.578125) – 1.859375 92(01) , 42(0.75)+0.25(+17,1875-0.578123) -0.9375 error. E ly, l Yent boron (u2 Inent Error O I TO 0 0 0 0.251.25 11.2441.0058 -0.1257-0.1588 0.033 0.50 5 1.4468 10.0531 +031251 -0.3952 10.082 0.75 1.7187511.54890.16970.5781 -0.7215101433 TOO 11.85937541.4686 10:39066-0.93754-1:143610.2061

RK4 Method koo hof (9 (4) 9 264) = 1 hos ng (Y, (t), G₂ (t)) 2-0.5 - Kia hf (9, (+) + 60/2 " (t) + 10/2) = 2 dah g (3, (t) + Kol, I 9₂ Ct) +10/2) 2-1 K2 2 h +ly, tt) tki, " (t) + 4/2) = +0.5 thang (%, (t) + K/ Y, (t) + 2/2) 201.25 Kz z hf 19, (t) tk2, 9, (t) + 12/2-1 Izah g (Y, (t) + 42, Y₂ (+)1/2) = -2 t() a go City,(0)416 (ko+2K,+2K+K3)2 15 Yz (1) 2 Y₂ (0) t/6 (Hot 21, +212 +d3) 221.1666 Eroron I sent Gorror 19 gent le Totögeron 10 1 ss 1-416810.0532|1564-7143640.02,36 lesser i heven in Eron in RK4 in larger step size.