Question

Question 2 (20 marks) In natural gas transmission, gas hydrates are a crystalline substance that can block pipelines, if they are allowed to form. A simplified model for the rate of change of the number of gas hydrate particles can be represented by a first order ordinary differential equation of the following form: dy Kty=0 dt Given that K = 1 and that y(0) = 100 compute the value of y(1), using the Adams-Bashforth Method with a step size of h = 0.2. In order to generate the first four points, use the 2nd order Runge-Kutta method.

Answer 1

Given dy - kty = 0 dt given R=1, at tro, y(0) = 100 . By using and order Rk method. We have to find the values ya at troi2, 0.4, '0.6, 018. Substititute Kal in equation : dy - ty=0; let = f (ty). dt . dy ty: ::. f(+14): ty. formula for finding y values by using and Order Runge-Kutta method : Yidt = i + [ Kitzki Jan. where his step size haoia. : kis f ( titah, Yi+ n) K2 = f ( ti + 2 h, Yi+sekin). i at tao Ki+262 x02 . I kit2k2 yotl L 3 fcs Scanned with or 2 = The CamScanner

Kif (ti+ 3/4h, Yi+3/4 h).. at t=0, Yo = 100 (Given). Ks = f (to+ 314x012, Yo + 3/4 2012) , ki= p < 0+3/4x012, 100+ 314.40.2 ) = +1.0.15, 100.15 ) if(ty)= ty.. if (0.15, 100.15 ) = 0.15X100115 = 15:0225 Kız 15.0225. K2 = f (ti+ 34hi yit 314 kchl.. -- Ef ( 0 +34X0-2, 100+ 3/4 X 15:0225 XOR) de z p10115, 102.2533) f(tig) a ty! f(0.15, 102. 2533) = 15.337995 --- K2 = 15.337995: Yoia a yot. f. kr + 2k2 ] +0.12 3 100 + p 15.02.25 +2 (15.33 7995) 7 Youn = 103.046 OY. (02) = 103.046 Scanned with CamScanner

at t=012, 41012) - You2 7.1031046 . . from the second order rk method. Yory = you + f Kit2k2 ] x02 . | k = f ( toa + 3/4 b, Loa + 3/4 5 ). = f( 012+ 3/4X012, 103.046+ 3/4 *012) 0135X103.196 ki si f ( 0135, 103.196) Eki Ki 3.36.1186. K2 = + (torz + 314h, Yoi2+3/4 *Kıx h) if (012+3/4X0-2, 103:046 +314.* 361118 6 X012) Ke = f (0135, 108,46379) K2 : 0135X 108146379.. K2 = 37,9623 you = youa + [ 36.1186 +2 * 37.9623] XOR 3 => your = Y(04) = 746 you 1031046 * 74695 70.4 = 4 (04)= 1lois155 at t=014 ,4001+) = 110.5155 second order rk method we find y From the value at Ecs Scanned with CamScanner t2016 l ede

Yous a Yout + Ritzkr J x 012 ki = f ( to14 + 314h, 4014 +314h). ki = f (014 + 3/4 X012, 11015155+0.75X0:2) kis # (0:55, 110.6655) → ki: 0155*1106655. ki = 60866 K2 = f (t04+ 3/4 h, Your + 344 *KIX h). K2 = f ( 0.4 +0.75X0.2, 1105155 +0.75X60,866X012) Kz: (0.55, 119.6454) K2 = 0.55X119.645+ K2 = 65, 80497. 1 bo 60866+ 2X65.80 4977 906 - 110:5155 + 3 1105155 4128317 12313472 . : at t206, 46016) = 123,3472 . From the second order Rk method we find y values. at taois : Yous = Youo+ s Kotzika] xood. ikaf (toir + 3/4 h, Yoig'+ 3/4 h) if (016+ 3/4 X012, 123,3472 + 3/4 *0.2) = f(0.75,123,4972) ki = 0.75% (2314972 = 9816229 . es Scanned with

ki = 92.6229 Kz = f ( tois+3/4 xh, yoxs+ 344 *KIX.6.) z f (016+314 X012, 123-3472 + 5/4 926219xo. Ef 10175, 137,2406)... . 0175x137.2406. = 1029304.. 9216229+2 (102.93 yous = 123.3472 .+ 1200 Yoox.= (43,2461 Now, till now find the values on y up to tois. Now we are finding y value at tal by... using Adam bosh forth method. Formula for of Adam-bash forth method Yiti = 99 + bis 55.fi - 59 fir +37f9-2-961-3:] Here we are calculating sth value of y So, let y) = yg and it,&=ts to8 = tq tor6 = t3 I . 4(0.8) = 44 y (0.6) = 43 Ý (0.4) = 42 y.(012) = y (0) = yo. to14 = tz to.zati to = to. . CS Scanned with CamScanner

] Prediction : YsPt you that [55f4- 59 fg + 39 fiz caf, fp = f (ta, ya) fzať (t3,83): fq = f (tz, Ys) 'fy = f(ty 912 we have the values ti=0,2 tr 2014 t3 = 0.6 t4 2018 40 = 0,4,-1031046, 42 = 110.5155, 43 = 123-347 2. += 14 3.346 4 = f (0:8,143.2461) = 114.59688. fz = + (0:6,12363472) = 74.00832 to = f (0.4, 11015155) = 442062 * sf (0:2, (03046) 20160921 012 – 5s (114.59689) - 59 (7400832 ) 143,2461+ 95 43.2461+ 24. +37 ( 44:2062) - 9 (2016092)] = 143.2461+ 02 ( 3386.48+49 ) ysP 1761 466804 1. fsP = f ( to, ysP) = f(!, 196.46680+1) its PE 1X171.4668041 fsPa 171.4668641 Scanned with Scanner

P Correction yiti = Yi + 5 (litr + 19 fi - 5 fi-, + ¥i-2) yer yn + h (født 19 fe - 5f3+ f2).. #s!= 1711 4668041, f4 ="114,59688 +z = 74.00832, f2 = 44.2062 , 94 = 143-2461 Ys = 1432461 + bid (171.4 668041 + 19 (444159688) -5(7400832) + 4412062) = 14312 461 + x 2022,972) 24 Y5 : 16011042 Here we are used two methods to find the value of Ys, One is prediction method and correction method the values we get the corrector method is our required value mwe get in. : Ys = Y(1) = 160.1042 L a 7(1) = 160.1042 Scanned with CamScanner