Question

3. Consider the initial value problem y(t) = y, y(0) = 1. a. Write down (i.e., write the formula which describes one step, Yn+1 = yn + ...) the second order Taylor method with step size h for this initial value problem. b. Write down the time stepping formula Yn+1 = Yn +... for the modified Euler method 9n+1 := yn + hf(en +3.29 + s(tn, yn)), for this initial value problem. c. What is the difference between the two methods in one step? d. The difference between the second order Taylor method and the modified Euler method in one step of size h is (hº). How big is p?

Answer 1

Solution:

Given that: The initial value of y'et) - y YCO)=1 forse: The second order Taylor method with step size h for this initial value problem. Y(tn) = yct) th h[sẽ tạº) Intl = ynth (ynt hy'(n) y'n=yn, yn=2Yn ". Into Ynth cyn + ₃ (2Yn)) There fore Inti = Ynthlynt hyn) t.. for ob The time stepping formula Intl =Ynt for the modified Euler method. Yato:= Yntnf (tat 22 yn + h / f (tn Yn)) for this initial value problem. Yatı : Yn + ht [f Ctnt ), yn 4 4 Ctr-un] f (ta.YA) = flent şi, ynt &fcta. Yn] + [unt be corra] Therefore Ynt i= Yath n[Cynt ons] Intl = Ynth Cynthy + hyn) ..Yu+iz Yuth Cyöt hy å ta ay )

for :- The difference between the two methods in one steps, Y, = Yoth (yo thyo) 4, = Yoth (yöthyo the guy. to o Therefore Y, = thith) y = 1th (1th hijo Yo=1 Difference : A4, = Y Yo Therefore Ay,= (1th (1th )) - (Ithith) - Ay, = x+y + hit h 22 - x - hika => Ay, = ht =) ay = no -K 1 => Ay, = (4) cho ... Ay, - úch3 fords:- The difference between the second order Taylor method & the modified Euler method step of size his och on one his oche difference о си*) : o che) Here P3 1. P=3