Question

For the following segments, consider the first order initial value problem : () =(t)-tº+1, (0) = 0.5 1. Solve this problem using RK4 with h=0.5, where from t=0 to t= 2 with step size h=0.5, it takes 4 steps: (stepo, to =0), tı = 0.5, t2 = 1, t3 = 1.5, 14 = 2.

Answer 1

Solution : Given that Consider ut) = u() - +1 – yo)=0.5 and n=0.5 the equation 1 i die-y-e+ dy. f(ty) f(t,y) = y-t+1 - © too, yo = y) =0.5 y(toth) = y, - yott (ki +262 +263 +ku) . Where ki=hf (to, yo) K2 = hf (to the yok ) k3 = hf (toth, yo t k 2 ) kn = hf (toth, yo+k3) ki = hf (to yo) : F(to yo) - Yo-tôt! f(0,0.5) = 0.5-603+1 f(0,0.5) - 15 ki=h f(logo) = 0.5 1.5 |k, = 0.75]

༥ - v(༩༦༤༠༠ ༌ ， 90+ ) （༤༠༧ ༧ , vo+) - (༢༠༠,༠༦ ༠༣) ： {( ༠༧s,0915) - （༠325)- （os) + | IES2900 - St&Q= ༽ • ར25 དར༔ – »(༩ （co, ， 9。 + པར་) = 5.5 （ །: 9125 [k，-༠༢3629 kའུ = hF (cosn ， 9o+k ) ༼ (༠༥༠ ༠:5 + ༠༠0༦) { (༠༤ཏ༠. qs3) - (6.45)-(os)-) - （3655-6.5625+། =།:825 3-5.5ཌ།:\25 [73= 0:036

③ ku = hf (toth, Sotka) f(to th, Yo +k3) = f(0 +0.5, 0.5+0.936) =f(0.5,1.436) : f(0.5,1.436) = 1.436 - (053 +1 = 1.4436-0.25 +1 = 2.186 ku = hf (toth, Yo tk3). = -f(0.5, 1-436) = 0.5 X 2.186 = 1.093 [k4 = 1.093. Y, = yo+ (ks +262 + 2k3+ km) +5+2(0.906) +2 (0.936) + 1.093) = 0.5+4 (0.75 +1.812 +1.872 +1.093) = 0.5+ < (5.527) - 0.5 +0.921 1y, = 1.421)

Now ti=0.5 and y(tid=4, ..^, = 1.421 find yltz)=42 Y2=9, +7 / (k, +Zk2 +2k3 tku) kishf(t, 19) f(0.5, 1.421) = y, -t, ²+1 = 1421 - (0.5)+1 = 1.421 -0.25+ 1 = 2.171 K, = n c f(0.5, 1.421) = 0.5 X2-771 = 1.0855 TK, 1.0855 K2 = hf (t) + 4 , Y, +K 2 ) f(0.5+ ors1:421 +1.0955) > F(0.75,1.964) = 1.964-( 1.964 - 6562+1 = 2.402

K2 = 0.5% 2-402 kz = 1. 200 k3= hf (tith, 9, tkz) *(0.5 +0,, 1421 +1-20!) = f(0.75 , 2.0215) F(0.75,2.0215) = 2.0215 -(0.75 +1 = 2.0215 -0.5625 +1 = 2.459 K3 = 0.5% 2.459 (k3= 1:23] ku= hfC Eith , Y, tka) f(0.5+0.5, 1.421+ 1-23), F (1,2-651) f(1,2.651) = 20651-C15+1 = 2.651-1+1 = 2.651 K4 = 0.5% 2.651 k4= 1.32551 42= 9(1) = 4, + 6 (1.0855+2x11.201) +2(1.23)+1-3255) - 1421 +1.212 Y₂ = y(i)= 2.633.

9324(1-5) =42+ = (k)+ 2xz+ 2x3 +ku] © y 2=2.633 (3= 1.5 62 = 1 ki = hf (tz, B2) f(1,2.633) = 2:633-1 +1 = 2.633 F (112.633) = 2.633 k, = 0.5X2.633 = 1316 k, = 633 k2 = nF (+24th / Yetki) f(14906, 2-63371.326) F (1.25, 3.291) = 3.291 - (1.25)+1 = 3.291-1.562+1 = 2.729 K2=0.5X2.73 = 1.3645 [k2= 1.364) k3 ehfQt2 th , 9z+ k_) *(1+015, 2-633 +1:3445) = f(1.25, 3.315) (1.25,3.315) - 3.315 - = 3.315 - 1.562+1 = 2.753 K3=0.5x2.753 TK3=1.376

Ku= hff tath, Y2 tk3) f(1+0.5, 2.633 +1.376) F (1.5, 4.009) = 4.009 -(1.53ti = 6:009 -2.25+1 - 5.009-2.25 = 2.759 K4 = 0.5X2.759 TK4 = 1.379) 42 = 2.633+ě (1.316+ 2(1-365) +2(1.376) +1.399) = 2.633+ $(1.316+2.73 +2.752 + 1.399) = 2.633+6 (8177) = 2.633 + 1-363 Ty3 = 3.9961 use u 43=3.996 t3=1.5 Yu = 43+ < (ko+2K2 + 2x3+ky)

Kız hf (63,43) f(1-5, 3.996) = 3.996-Clstil - 3.996-2-25+1 = 2.746 ki = 6.5X2.746 ki = 1.373 kz = hf (t3th 2 , 93 +4 + 2) + (-54% 344 642) +(1-75, 4-693) = 1.683-C1-75)*+, = 4.693- 3.063+1 = 2.62. K2=0.5x262 1. Ke = 1,31] k3= hf (63+h 4, 93 +K 2 ) F15+, 3-9967 1) f(1-75, 4.651) = 4-651-(1.753+1 = 4:651-3:063+1 2 = 2.588 K3 -0.5 X 2.588 [kg = 1.294]

ku – hf(ts +, Уз+ks) f(1.5 +0.5, 3.996+ 1.294) F(2,5-29) = 5.29-471 = 2:29 K4 = 6.5x2-29 |k4= 1.145 Yu = 9(2) - 3.996+ Ś (1-373+ 2(1.31) +2(1.29u)+°45) Yw= 3.996+ + (1.373+2.62+ 2.589 +l·lus) yu= 3.996+7(7926) = 3.9967 1287 yu= 5.283. y (0) = 0.5 y (0.5) - 1.421 y (1) = 2.633 y (1-5) = 3.996 y (2) = 5:293 Please give me rateing for this answer, Thank You.