Question

3. Euler's Method (a) Use Euler's Method with step size At = 1 to approximate values of y(2),3(3), 3(1) for the function y(t) that is a solution to the initial value problem y = 12 - y(1) = 3 (b) Use Euler's Method with step size At = 1/2 to approximate y(6) for the function y(t) that is a solution to the initial value problem y = 4y (3) (c) Use Euler's Method with step size At = 1 to approximate y(0) for the function y(t) that is a solution to the initial value problem y = 21-ty, (2) = 2 Hint: Your initial condition is after the point you want to estimate to - you'll have to step backwards. (d) Use Euler's Method with step size At = 2 to approximate y(8) for the function y(e) that is a solution to the initial value problem y = 4y, y(0) = 0 2 MATH 12 Worksheet 10: Direction Fields and Euler's Method S2019 (e) Consider the initial value problem v=y-2,0) = 4 Do two steps of Euler's method with each of the following step sizes: At = 1, A4 = 2, At = 1/2. Compare these results to a stability analysis of equilibrium solutionsfor the differential equation, and discuss the consequences of approximating with these different step sizes

Answer 1

300) we have yl=f7-4, At=1 Y(1) = 3 then we have to find y(2), y (3), Y14). SOLUTION = B (ho Zulh's rulo ( 0 = h (54))) Yuri = In thif (thiyn) , where tnt = futh STEP-J. - t = toth Y(+1) = 4(2) = yoth. f (to, yo) 3+1(-7.0) - 1 =7y (2) = - ) +1 = 2 3+ 1. f(1,3) = 9(2) = = 4, (say) STEP-III- Ez = tith = 2+1 = 3 y (H2) = (3) = 4, tch-f(+1,91) = 1 + 1. f(2, 1) 1 +1: (3) = 4 y (3)=4=Y (say) STEP-W! - f3= fetch = 3+1 = 4 Now y (4) = Y2+ hf (+2, Yz) = u + f(3,4) Yly) u + 5 =9 914)= 9 Answer.

Then y(6) = 2. SOLUTION! - we knew that Euler's method 3(b) yl = 44, gt = h = /2 , Y (3) = 1/4 STED-II- 712 STEP-II!- Intl = Yutchiffen, Yn) where tuti= futh to = toth = 3 + 2 Y(+1) = 4(7/2) = Yoth f ( to, Yo ) = 4 + 1/2 (1) = 0.75 y (7/2) = 0.75 =9, (say). t₂ = titch = 7/2+1/2 = 4 y (4)= Y,+ h f ( t Y,) = 0.75 + £ fl 7/2 10.750 = 0.75+{(3) = 2:25 y (4)=2.25 2 lsay) STEP-111- t3= tz tch +3= tz+ch = 4+2 = 9/ Y2+chf (4219) 2.25+ $$14,2-25) 2.25+}(9) = 6.75 = y(9/2) = 6.75 = 3 15uy) STEP-VL tu= t3 tch= 9+2 = 5 915) Yz + hf (63,93) 6.75 +5f($16.75) = 6.75+ (27) = 20:25 y (92) = y(5) = 20.25 = Yulsay)

STEP-I! - ty = futh = 5 + 1 / 2 y (1/1) Yuthf(ty, Yu) } = 20.25 + $(5,20:25) = 20-25 +2181) = 60.75 y (11/2) = 60.75 = Yglsay) STEP-V1- €6= 65th = 14 + 1/2 = 6 916) = 45+th $(t5,75) = 60.75 + £ $(4960-75) = 60.75 +1 (743) y 167= = 182.25 Answer.

300)! - y = 44, y 10) = 0, At= ch=2, Y (8) = ? SOLUTION! - we know that, Euler's farmula. Intl = Ynt hf (Insyn), where tnt1 = tntch. Here f(ty) = 4y ISTEP-I fi = to th=0+2=2 y (2) = y + hiflto, yo) = oth f10,0) - 0 +2.00 I (say) = 4122=0 STEP-II - t2 = fitch = 2+2=4 y (4)= y + hf (429) = 0 +210) = 0 - 414)=0 o+af (2,0) = Y(say) STEP- III E3 = fatch = = 4+2=6 y (6)=Yz + hf (42,92) = 0+2 f(4,0) 6 +2.0-0 = y (6) = 0 - Yulsay) STEP-NI ty = t3th = 6+2=8 418) = Yu+ch f (t3, 93) = 0 +2616,0) = 0+ 2.0 =0 => 118) = 0 Answer.

SOLUTIONI - Euler's formula is 913(1) yl=24-ty = f(4,4), At=ch=1, y (0) = a (say) = ? andy (21=2. ( Criven). Into = yn + h flfni Yn), where tuti = futh. STEP-I:- tia toth = 0+1=1 911) = yotch f ( to, Yo) = atlifco, a) yll) a + 1.20) =a = y, suy) STEP-II - Ez = fitch=1+1=2 y (2) = Yitchf (6,,Y,) = atl.fli, al = a + (2-a) 2Y (2) = 2 which is independent froma. So, whatever be y(0), we will always get y (2)=2. So, yeo) =a is arbitrary,