Question

Question 3 døy Not yet answered Marked out of 2.0000 P Flag question Consider the following Ordinary Differential Equation (ODE) for function y(x) on interval [0, 1] dy dy + (-8.6) + 14.03 dx3 dx2 dx +(-2.47) + y(x) = 3.762 with the following initial conditions at point x = 0: dy y = 4.862, = 15.4696 = 77.4217 dx dx? Introducting notations dy dydy dx dx dx2 convert the ODE to the system of three first-order ODEs for functions 1, 2, in the form: =(x, y, 92y) dx = /2(x, y1. 2. y) dy dy2 dx dy dx +63(x,, ) Apply Euler's method with h = 0.5 to solve this system numerically. Fill in the blank spaces in the following table. Round up your answers to 4 decimals. y3 o 0.5 1.0

Answer 1

dyz = (8.6) 7 - 14:03) g +2 +41)), Here our dy system is, dx = y dge da cha ge=dy a ax +6.762) 2062-6 Equation o s obtained from the given ODE after putting G day da² Here, Noo We can take the value of f, at no=0 as yo e To 4.862 stmularly, - 15.4696 Uso = 77.4817. 720

h=0.5. -the gino lAfrientiel grandiosa dy = f(ny) then in Luler method we use the formalar Int) – + bxf (d. ) evaluate the valence of y corresposa ding to a partienas We can start from Ignation @ following iterations can be nsed. don - Y, Hvor eq0) ca I= Go + bxfi (As, Ho, 9 201930) 4.862+ 0.58 Yo [ Here - 4.862 80.5 x 15.4696 to = 4 S 12.5968 my value of y, at No=0.5 is 12.59

Then consider second equation dy an Jest bx fz (vo, To G. 19) (1.20 Here, ₂ - G equation, 9x) = 15.4696+0-5X 17.40 11 51.18 045 51.1805 in value of g at n = 0.5 is, 51018045 Consider third dys (8.6); - (4:03)47 2.47)44 46.762).2014) : Igi = 30th & f (no, o 19) Here £3 € 18.6) 43 (14.03 ) 2 + 2:41)4, +3.762)206-27 ca

re value of g at no is 309.702336- 6. 9. = 17.4214 +0:5[8.6(79.4917) - (1903) (15:4696746-47)4-863) +8.762).8840] - (97640211) +0.5 64 64.55997) =3090701336230901013 yz Consider the second iteration, there, N= Not 0.55 0.5 9 12 = Guthx filan y Gall Isl) - 12.5968 + 0.5 (5 40 18045 39.6 87025 r 39.6870 thx fa (1) I J 211 Ys) - 54.18 245 40.5 (309.701336) 209.031118 1209.0311 Gith && (81 191, 92, 1931) Yaa Y21 732

here, 4th = 309.701336 + 0.5 18.6 X30 91401336) (14. 03 X5 4.18015) + (241) (6.596874 8.162)x64 - 1097. 10439 436 63 2,1297-7044 Conssder the third iteration 2 913 - Sat hefi (x2, 4,2 1922, 932) = 39.681025 +0-5x209-03118 1440 202584 144.2026 9 a3 = Yed + bxf, (2, 9, 21 J22) 932) -209.031118 +0.5 x1279-704394306 - 848088334715 32 N 84, 8.8834 Iz = 32 thx fi (22) 2172 21832) 12274. 104393063+ [8.6*1279. 70.439

= 1219-70 43930634 (8.61214.-104943063 - 14.03209.031118+ C-47)x89. 684005) + (3. 762)x -6] 5362.2777141716 -1536 २,2717 47 42 13 10.5968 51. 1805309. 7013 0.5 24.6870 २०१.03111297.78.45 1 14410026848.8834536212797