NEW TOPIC ADDED FROM ACADEMIC SESSION [2022-23] THIRD SEMESTER [B.TECH] COMPUTATIONAL METHODS [ES-201] UNIT -1 Q.1 Give geometrical interpretation of Newton Raphson Method. (2015) ‘Ans. Let x, be a point near the root « of equation fix) = 0, then tangent at Alx,, flx,)) is y ALK fo) y-fix,) = 20x) (x — x9) 1%) Tecutsxcaxisate = x5 = fs ») which is I approximation to root u. If, corresponds Xo) to x. on the curve, then tangent at A, will cut x-axis at x,, nearer to a and is therefore 1 approximate to root «. Repeating this process, we approach the root a quite rapidly. Honee the method consists in replacing the part of the curve between A and x-axis by the means of the tangent to the curve at A, Q.2 Using Bisection method find the value of negative root of the equation x!-x +11 = 0 upto 2 decimal places. (2015) Ans. = xi-x+11=0 f(-2) = -8+24+11=5 -274+3411=-13 fi-3) Root lies between — 2 and ~ 3. x, = fix,) = — 15.625 + 2.5 + 11 = ~ 2.125 (-ve) Root lies between — 2 and ~ 2.5 x = T2te25) - 9.95 - 2 fix,) = 1.85 (+ve) Root lies between ~ 2.25 and - 2.5 x = 2225+2-5) -_o 975 fix,) = — 0.02 (-ve)2-2021 root lies between ~ 2.25 root lies between ~ 2. 12a % fix) = % fix) = fix) = ‘Third Semester, ComP and - 2.375 wutational Methods 2,95 + 2.375) 72.25 2ST 2 2.312 fix,) = 0.953 (+ve) nd - 2.375 2.343 9.912 + (2.375) =2.312+ (2.575) 2 0.4807 (+ve) root lies between ~ 2.343 and - 2.375 343+ (-2.375) ae 2 0.2314 (+ve) 359 _ root lies between ~ 2.359 and ~ 2.375 -2.350+ 288) arse 0.1054 (+¥e) * root lies between ~ 2.367 and -2.375 Required root is - 2.37. Ans. fix) = fix) = £10) = root lies between 0 and 1 Again £(0.5) = £07) = Let x, = By Newton's formula 2.367 + (-2.375) 2 -2371 Q3 Use Newton Rephson method to solve the equation 3x-cos x1 = 0 correct to four decimal places. 2015) 3x — cos x -1 3+ sin x = 2 (-ve), f (1) = 1.4897 (+ve) 0.977 (-ve) 0.33 515 (+ve) thus root lies between 0.6 and 0.7 06 a Let first approximation be x, = 0.6 x,sinx, +cosx, +1 = “fink te0sx, +1 S+sinx, LP. University-[B.Tech|-Akash Books 2021-3 = 0.6s8in0.6+ 0080.6+1 _ 9 gar 37 5in0.6 a xysinx + 608%, +1 ‘B+sinx, = x, = 0.6071 root is 0.6071. Qu Evaluate /j2 to three places of decimals by Newton Raphson method. 2017) Ans. Letx = Vi2 >x*-12=0 fix) = x1-12=0 “a 13) = -3,f14) ‘Thus root lies between 3 and 4 Let x, =3.5 Iterative formula = < Ai) Putting x, ‘Thus root is 3.4641 Q.5 Using Newton Raphson method, find a root of the equation xin x + cos x= 0 which is near x = x correct to 3 decimal places. (2018) ‘Ans. Let f(x) = x sin x + cos x Let 4( ) = Reve f(a) = -LGve) root lies between = and x Let xen By iterative formula x,,. = ¥,‘Third Semester, Computational Methods = Hpe0SKy Eq SiN Ey OE nk , CO8%, ee (2.8238) cos(2.8233) - 2.8233sin(2.8233) cos(2.8233) 2,8293c0s(2.8233) x, = 2.7986 55 .7986)° cos(2.7986) - 2.7986sin(2.7986) — cos 2.7986 2.7986 cos(2.7986) = 27983 Required root is 2.798. Q.6. Use Bisection method to find a root of the equation x* - 3x ~ 5 = 0. (2017) Ans. fix) = x-3x-5 f(2) = 8-6-5=-3(-ve) f(3) = 27-9-5= Root lies between 2 and 3 Let ees y= 5F-3x25-5 3.125 (+ve) fix) Root lies between 2 and 2.5 2425 =a! A 25 2.253% 25-5 = 0.359 (-ve) Root lies between 2.25 and 2.5 R= fay = 225+ a 2 25 2375 fix) = (2.9753 x 2375-5 = 12715 (sve) Root lies between 2.25 and 2.375 = 225423 x +2575 «9.3125 2 = 0.0692 (-ve) 35 and 2.2815 2.2785 + 2.2813 _ 5 0974 Root lies be fix,) = 2.27749-3 x 22774-5 = 0.0203 (ve) Root lies between 2.2774 and 2 2.2774 +2.2813 _ 9 794 2 2794? 3 x 2.2794 —5 fix,) = 0.0048 (+ve) Root lies between 2.2774 and 2.2794 2.2774 + 2.2794 _» ong4 2 = 2.9784? -3 x 2.2784-5 = -0,0078 (ve) Root lies between 2.2784 and 2.2794 2.2784 2 2794 9 o7a9 ) = 2.2789? -3 x 22789-5 =—ve fix, Root lies between 2.2784 and 2.2789| Methods 62021 ‘Third Semester, Computational x, = 22184+22789 _ 9 9787 3 = fix,) = -ve Root lies between 2.784 and 2.2787 2.2784 + 2.2787 _ 9 o786 2 fix,) = -ve Root is 2.278. Q7. Find the Newton Raphson iterative method to find the pth root of a positive number N and hence find the cube root of 17. (2017) Ans. Let x= NM = x-N=0 ed fix) = *-N=0 =. 1) = gx Iterative formula Four N=17,q 2) 12) (3) = 3-17 (+ve) Let x, = 2.5 by (2), we get 22.59 +17 oxo = 2.5733 2x (2.5733)* +17 3 x(2.5733) Med 2x(2.5713)' +17 3*(25713" = 2.57128 3x(2.57128) mare Root is 2.5712, LP. University-1B-Tech]-Akash Books 2021-7 Q8. Define absolute, relative and percentage errors. ‘Ans. Absolute Error: If is the true value of a quantity and X’ is its approximate value, then | X—X'| is called the absolute error F,, Relative Error: The relative error is defined by Percentage Error: The percentage error is x-x'| mal Q.9. Find the absolute error and relative error in /6 + /7 + 1B correct to4 significant digits. ao) Ans. Let Let = 0.000668 | X-X’ |_ 0.000668 |x |" 792366 Q10, Explain Newton-Raphson method with convergence. Find the positive root of x‘ - x = 10 correct to three decimal places. Using Newton- Raphson method. 20) ‘Ans. Newton Raphson method: Let x, be an approximate root of the equation fix) = 0. Ifx, =x, + h be the exact root, then flx,) = 0. Expanding fix, + h) by Taylor's series fix) + bfx) Pris ate ° Since h is small, neglecting h? and higher powers of, we get fix,) +f") = 0 £(x9) F(x) A closer approximation to the root is given by > h x, £9) a; ig ae milarly starting with x,,a still better approximation x, is given by «, £00) 1 ta) ~ £45), 9 =0,1,2,3,.. In general, age ee 7 fs f'x,)hods Sie ‘Third Gomester, Computational Met Which in known as Nowton-Raphson formula see Convergence of Newton-ftuphson Method peas tity on no that =e Suppose x, diffors from the root a by w small quan ae ond x eo Oy 0.0668) O68) 1.8665 ‘Then the general equation becomes fi f law £00) = 0} are already present in the statement of ‘nt errors. ‘These errors occur either due ins of calculators, mathematical e from the process of rounding off the ding errors, Also called as procedual by changing the calculation 1 each step and rounding off renee 4 » X (say) is truncated to =-10 X’ (say) then truncation error = X ~X’. ference between the true values of @ X is the true value of a quantity and X’ is its 's called the absolute error, denoted by ¢, root is nearer to 2 XX! | 251 = wee Q19. Use golden section search to find the value of x that minimizes fix) = x! fx" + 60x"— 70x in the range {0, 2]. Locate this oro. Ans. After N stage the range (0, 21s Choose N so that ta la, bJ = 10, 1.236] 1s oe . 1-25 e =—*——__ >. **, a * % sb by = —————————— RaaeT Iteration 2: We choose b, to coincide with a, and f need only be evaluated at one ae nn LP. University 1B Techi-Alcenh Bockes mon-a6 a, = a+ pb,-0) = 04721 fia) = 212 fb) = fa) =-2636 fia,), 50 uncertainity level is reduced to la, b,) = 10.4721, 1.236) i i=. t 2 3-235, Now, fib.) < Iteration 3: We set 2, = a, and campate b, b, = a,+(-p)(b,-2) = 09443 =-2436y Semon ‘Third Semester, Computational Methots UNIT - 5 - 2B Compute tog, 224.5 trom the table be Papers gs x: 19 7S me 8H be | ; jog: 2G tO 251% «245119 Ans. x fix) Alix) ofl) 4 ole © Stempoon's one third rete gwen thet ¢ = 272, AS and compare it with the actus vaben. ons 7 OA 4245 ~ #21) 7 00012 : Q21 Using simpson's one third rule evaluate xdx Upto 4 deci place in 10 intervals, i ina oy, a om ay ah 0B ae, a Tf20 07910 =y; * %, h gi%ot Yio) Ms 45+ 96+ 9, +95) = lee 24 y@) 2 . +20, 49,4949) (i) Wo (f(x)Pes strand Sesestr, Computational Methods Q.24 Pind the number of men getting wages betwee the following data Ans Table is rearranged as Difference table is LP. University-(B.Tech|-Akash Books 2021-19 wages (x) No.ofmen ay aly YY tess than ” terpolating polynomial for (0, 2), (1, $) 2, 12) and (5, 147) polation formula. (2015) verge EOE DE=2 347 6-06-62 1862 aye By» GaP he B 6 60shird Semester, Computational Methods ; ; 3 2 4.172=10)+ 2 (x9 722 a fix) = tats 4115-10 +4 (29-72? +105) 2 Bava, 3 _ gx? +5x)+ (0? -3x7 + = x3 -6x" +52)+ 55 2x) fi) = A(x? - 82" + 17-10) + 15x" -727 + 19) ~20(x3 — 6x? + 5x) + 49(x° -3x7 + 22) alessio es et 2) 20 40x5-100x* +80x~40 A ee c 20 fla) = Qe 5x2 dx +2 a ASL Q.26 Evaluate (he by using simpson’s > rule taking 4 = 1 a 4 3 , 1 aii y Usit fe h = 1 where h is interval of di ‘ by using Simpson's = rule taking h = = al of differencing, 2015) = 1a-05)+4(18+0.64)+2. = 0.785392 1 fo) = 1, -16 ext x 0 6 36 46 56 1 fx) 1 3687 910 4% 913 3661 12 Yo y % y By Simpson's Sule [faite = iy, T B+ 96) 80 +¥2 +94 +95)+2y5] 2021-21 a factorial notation hence show that Q.27 Express y = 2x°- 3x" + 3x-10 (2016) ey = 12, Ans. y = 2x! — 3x7 + 3x— 10 y= Ox(x~ 1) (x-2) + Axx- 1) +Bx+C 2x! = 3x? + Bx — 10 = Axx — 1) (X= 2) + Axx 1) + Ba + C Forx=0 = -10=C Forx=1 => B+C=-8 = B=2 Forx=2 = 2A+2B+C=0 = A=3 y= 2ixl? + Six}? + 2ix] - 10 Now, Ay = Glx}! + Glx) +2 aty = 12Ix] +6 sty = 12. Q.28 From the following table, estimate the number of students who ybtained marks between 40 and 45. (2016) Marks 30-40 | 40-50 | _50- 60-70 |_70-80 No.of Students [31 42 En 35, EN Ans. Tal earranged as ‘Marks obtained No. of students (C.F.) less than 40 a1 less than 50 73 less than 60 125 than 70 159 less than 80 190 = ame mar] [oy] oy ey o aT ica]Computational Methods 22-2021 ‘Third Semester, {We calculate f140) and f'45) (40) = 31 uD 42 as) = yt40) + nayt4o) + SE a’y(4o) = u(u-1)(u~2)(u-3) MU DIN= 2) A5y(40) + : ayaa 3! 4! 4 a=40,h=10 = 45240 205 a 5(0.5 - 1) (0. 2) 5-1) 0. -1)(0.5-2) qs) =31+05x42+ 5 05-D.9+ a af i oe x37+.. = 52 - 1.125 - 1.5625 - 14453125 = 47.867 = 48 Number of students between 40 and 45 = 48-31=17 29 Using Lagrange’s formula, find the form of the function f(x) gi ‘ io ) giver that. ou x o 2 3 6 fx) 659 705 729 304 Ans. fx) = F=DADE=B, gpg , E=-E-ME-8), 705 (0-2)(0-3)(0-6) 4&0) (x 2)(x-6) -0)8-2)3-6) (2-0)(2=3) 2-6) (x= 0)(x-2)(x-3) (6-0(6-2)6-3) “4 x 729+ 2 fix) = P=5Kx+6)(x-6), gag, x-6) oe , x05 OZ HB -6) gy , (x? 2K)(x-3) gy =I 6 = xx*-11x? +36x-36) (x° -9x? +18x) a6 659 + —— 5 «105 ~ (09 ~8x? + 124)81 4 OO = 5x" + = 6x) . 67 ~1318x 2 ee }14498x? —47448x +7448 + 6345x? = ~S7105x* +114210x - 5832x° + 46656x? 789984x + 804x" - 4020? + 4824 2 LP. University-[B.Tech|-Akash Books 2021-23 mx? + 1602x + 47448 72 fix Q30 Use Lagrange’s interpolation formula, express the function 3x‘+x+1___ asa sum of partial fractions. (2017) &_D&-2E-3) Ans. Let ee = (x1) 2Xx-3) Consider a(x) = Bx exe es for x= 1,2, Its ta ate v axt+x+ 1 Using lagrange ax-) &=2) Q31 Using simpson’s one third rule, evaluate [. ¢< Yenee find value value of log, 2. oe (2017) ‘Ans. We divide interval in 8 equal parts 1 Here = ——=fx) e y= py oh x 0 eee 3 0 4 8) sa 8 8 8 8 8 8 8 yo1 8 8) 8 aoe 9 5 L 3 19-7 ee y ye ee oe By Simpson's rd rule A h = Blo +ya) dn +9 t¥e ty) * Mat Y 0) jigs = H{(-4) (: 8 5)so(4+2-4)] = 0.699154 24 2, 13° 16, 5 3 7, 7 jes ite i Since {-% = flog,(+0]} = log, 2 ines Wir flog. +x], = log. log, 2 = 0.693154.e 24-2021 ‘Third Semester. Computational Methods .es of x and y- ving table gives the valu $32 The follo} | w2 [24 L_28 [ean | ae ee ; 2 oe [| _ os [ 134 [15.5 ais. z ding to y= 12, using Lagrange’s interpolation Find the value of x correspon y atin formula. ‘Ans. According to formula 12-62) x=2. 008 -28K5-4 tx 2 2.808 £0) = G9-912-2812-4 (x= LN 2.81x— 4.1 4.9)(x-6.2)6.8 B1=1.212.1-2.8N21-4.12.1-4.92.1-6.2) (x= 1.2) - 2.4.4.9) = 6, +8 -1.2N2.8-2.1128-4.112.8- (x=1.2hx - 21x 2.8Kx-4.9) 2021-25 mula find the value of y when x = (2018) + gre 1.2K4. 2.04.1 - 2.84.1 - a (x-1.2)x-2.D(x-2.8)x-4.1 “os 9-1.2149-2.04.9-2 : £12) = (12=1,212-2.812—4.1)12~ 4.9\12-6.2)x6.8 <3 14.4648 7 2.124.112 - 4.912 -6.2)*9.8 _(02-1.2)02-2.102-2. 12.6672 (-(02-1.212-2,102-2.8)12-4. 22.6262 302-1.2(02-2,1)2-2.8)02~4 628.1856 £12) = ~1610.84 + 15195.727 + 597.482 + 42850.637 ~ 30876.064 + 1721.473 7878.4 Q.33. Find the value of ft (€0s x~x) dx ia fp (ex PY Using trapezoidal rule with number of steps equal to 6, ee 2 on ‘Ans. Dividing the interval (0,1) nto 6 equal stops Here a5 asuh = = 100+50u = 218 = u = 236 By Newton's forward formula fla + uh) =y, + nay, + u(u-Wu=2) 43 ae A2y + Yin = 10.68 + 2.96 x.24 + 2981236-0, 2! +, (2:36)2.36- 112.362) a (0.15)Third Semester, Computational Methods 26-2021 (2.36y2.36- 112.96 -242.86-9)_-,07) e a (2,362.36 1X2.96— 202 8 Be 2 oc. + 0.0280 + 0.0000 + 0.9001 = 15.698 (2018) ‘Ans. Lagrange’s interpolation formula is (x= x Mx 2 —NE= 20) F¢5,) Teg HX — Ea Mo ~ X50 ~%a) fix)= (x= px — Xx HH x,) eas — NH =x ray Me= Hye N=) Ge Yo = 5), 209 -* lr) ea eXe ae ete 3y) reese Ngee ma) | Hata hT — 1X84 2K ay) (= Thx =11Yx-13 = fa) = 5 7x5- 1115 -19% BWR =1KR=ID, 509) (8X7 -11K7- 137-17) 452 pL BWR=Thx-1WE= 1D 9966 + (@3-5)a3-7)43-1)03-17) (x= x= 11hx- 13-17) _(x-5Xx-1x- fix) = xe bia 0.1302 0.6167 (¢ EL D= Tx =1IV 1) _ (5 5Xx-T(K= 1-17) 5.0417 6.1615, (_ X= 5M Dix 112-13) 1.8063 Atx=9 fig) = (9-79 -1149-1349-17) _(9-5\9-11X9- 139-17) 0.1302 0.8167 ¢ (9=5X9-7H9-18'9-17) _ (9-59-79 - 1119-17), 5.0417 ae 1» 1.2063 = 963.1029 + 913 4066 + 50.7765 + 20.7142 + 36-4315 119) = 562.66. Q.36. Evaluate (* 25 +e with its actual value. ow) interv to 6 parts each of width h = 1 Actual value Q87. Evaluate the integral [) "5 dx by using 2. Simpson's + rule 1. Trapezoidal = 3 3 3. Simpson 3 rule 4. Weddle rule and compare the results with its actual value. x +x Ans, Let yx) = : and Te fyde into five subinterval by taking h = 0.2 » [ s Jos xQ.40. Bvaluat the interval [0, p] in ‘Ans. We have fix) = x sin ‘third Semester, (-0.4X-0.4+1)(_9,0005) + Com} putational Methods Costoae) 2 = 1.18132 tetheintegral [x sinxdx sin subintervals. using Simpson’ (0.4\-0.4 yefx) | 0 Here y, = 0, ¥, = 1 Using Simpson's > ru [[xsinxdx = ras + 20.9073 +1 = 3.1454 QAI. Find the data: f= 5 7 | [te [150 _[_ 392 5202 Evaluate f(9) by using Newton's Ans. The dividend tabl (2019) [ Lt afi) sf | afte | styoo 1 392 m 1 n 1452 32 0 1 13 2366 0 Ww 5202 LP. University-1B.Tech|-Akash Books 2021-31 Using Newton's divided difference formula A(x) = flx,) + (x — 4) fl) + (x — Xe) (= x,) fl.) + (=x) =x) (X= x) B’flX,) + 1x) = 160 + (9.5)(121) + (9.5) (9 ~ 724) + (9 - 5X9 — 7)(9 — 112) = 810 Q.42. A second degree polynomial passes through the points (1, -1), (2,-1), (3, ) und (4, 5). Find the polynomial. (2019) Ans. The difference table is x yet | ay | 4 ay] r a o 2 Bi 2 2 ° 3 1 2 4 4 5 fe take x, = 1 s0 Pp aera er h 1 Using Newton's forward interpolation formula we get P(P-1(P-2) fx) = Yq + PAY + ae = -1+(x-1(0)+ 9 pe ) a = -1+0+(x-1)(x-2)+0 = t-ox+1 Q.43. Find the value of [*e* dx using Simpson's 1/3 and 3/8 rule by {ing the interval 0, 0.6] is suitable number of subintervals. (2019) [exo 0.3 o4 05 0.6 y=ftx) | 1.0000 [1.10517 [ 1.22140 | 1.34986 | 1.49182 | 1.64872 | 1.82212 , = 1.0000, y, = 1.10517, y, = 1.22140, y, = 1.94986, y, = 1.49182, y,= 1.64872, Using Simpson's one third ruleauird Semester, Computational Methods ya +¥e) foxa- yyy sye)e do t¥9t 9" 987% , 3 — 1.64872) + 2(1.22140 gpg) + 1.10517 + 124968 - 1.64 + 148169, = 92.0000 + 2 = 0.82212 By Simpeon’s three eight rule fierax = Bivorv0)+9% +9at74* 70)" 209 1.39517 + 122140)+(149182+ 164872)+ 201 34955 = 40.0) (1.0000 + 1.82212) + 30 3 2082211 Qu4. (a) Compute [sin xdx significant to five digits by Romi method. ‘Ans. Take h= =,%, successively and compute integral using Trapezoidal rule 4816 h = Buy, +yo)+ 2vi! give ¥2)* 2 = 2[1+2x0.7071)=0.94805 z ) When h = values of y = sin x x | 0 we | x4 | 308 | 2 y_ [0 ossz7| 07071 [ 05239} 1 Boia sa2%s joey Mh) = By, +y,)+ 27, +¥2 +90) gio +¥4)+ My + ¥2* Ys) = 211+ 20.8827 + 0.7071 + 0.9239)] = 0.98712 Gi) When h = 2 6 h cat) = Pity, +9, + 29,4324 Yat Yet Vet Met Id a) = Bling + 991+ 20,4944 Hat Yet Vee Tet H 7 {1+ 2(0.1951 + 0.3827 + 0.5556 + 0.7071 + 0.8315 + 0.9239 + 0.9608] 32 = 0.99680 First order extrapolated values are on Way 4-1 = 4£0.9871- 0.9480 _ 4 99915 4-1 Have, wa) = *1ca/4)= 1th 2) 4-1 _ 4%0.9968 - 0.98712 _ 1 999934 4-1 Its, w2, bia) = #1042, /4)~ 1 ih) 4-1 4x 1.000034 - 1.000152 4-1 = 0.99995 Value of integral = 1 Q45. Evaluate [} (x? +2x) dx by Gauss Quadrature formula. ‘Ans. Here a= 2,b=4 fix) = x8+2x , A) to [- 1, 1] by putting Let us transform the interval x= fla+b)+ (beady) x= He+d+4-aylComputational Methods ‘Third Semester, xa 8ty dx = dy T= fifeodx 6(y) = 113 + yy = fw By taking n = 2 in Guass formula b-ag 2 Rely) I 5 = T = IRoy,) + Ro where R, = 1, R= 1 y, = 0.57735, ,=-Ys 40) = 13+y) = (Gry? +(8+y).2=y'+ By +15 4y,) = yt + By, + 15 Now = $1y,) = (0.57735)? + 8(-0.57735) + 15 = 0.399933 — 4.6186024 + 15 = 10.7145 Hy) = (y+ Bley) + 15 = y,?- By, + By (a) = 1% 10.714531 + 1 x 19. = 30.6666 LP. University-1B Tech}-Akash Books 2021-35 UNIT - 101 QA6. Solve the following equations by Gauss Jordan method. (2017) Ox+y-4e = 1 6x + 9y-34 = 17 x-2y+2 = 9 ‘Ans. In matrix form, we have E ‘ ET. | Consider A 2 R, > R,-5R,,R, > R, +8R, fi 1/2 -2)/ 1/2] o 2 -15|| 5 2 1 1/2 -2][1/2) -[e 9 al at R,boehpeany 13 fints2. 29) tat ~ JO 13/2 13 || 29/2 0 0 ~19)/103/13. 2 RoR Em 1 0 -3]f 6/13 - [0 13/2 13 |} 29/2 0 0 -19]/103/13, 3 13 R, > R- Tg RRR + 9% [ 0 08 ae 0 13/2 0 || 29/2 0 0 -19]| 757/3810 Ol -195 / 247 pcets |e 0 0 12 757 1722. oy-195/247 ~ lo 1 of 29/38 0 0 1)[-757/722. Ersoy o29e-_ -187 = X87) 18" 782 thod. Q.47. Explain matrix inversion me! Let B be another matrix of the same org ‘Ans. Let A be e non-singular matrix. such that AB = BA=I Where I being 2 unit matrix of the same order. ‘There are two methodsL 1. Gauss elimination method: The order as the given matrix A and write it as Al ‘Now making simultaneous row operations on AL, we try to convert A into an upp: triangular matrix and then to a unit matrix, Ultimately when Ais transformed into unit matrix the adjacent matrix gives the inverse of A. 2. Gauss Jordan method: This is similar to the Gauss elimination method except that instead of frst converting A into upper triangular form, itis directly converted inta the unit matrix. Q.48. Explain UV factorization method and hence apply factorization method to solve the equations. x+y +2=9,x+2y+32=6,3x+y+2z=8. ‘Ans. UV Factorization method: This method is based on the fact that every square matrix. A can be expressed as the product of « lower triangular matrix and a8 upper triangular matrix, provided all the principal minors of matrix A are non-singular IFA= la, then method involves a-unit matrix of the same Jan a2 2 zag | +0 and soon. Sn Ss Mas Also such a factorization if it exists, is unique. Now consider the equations ah tah tum = by uh t+ 8o%+ 05x, = by LP. University-[B.Tech|-Akash Books yh + 8k + Oe = bs Which ean be written as AX = B. fe a2 Mis Where, A= sarag Lan as2 89 Let A=LU fa 0 0} if WhereL= |i) 1 o|and U= Lar 42 Od po ‘Then (1) becomes LUX = B Let (Xe ‘Then (3) becomes LY =B Which is equivalent to n= % Ins +9_ = Be =, Solving thes then we get Y. 4) becor uy, + U2 + UX, + Unt 's Um = Js From which x,, x, and x, can be found by back substitution. Now, we solve ox+3y+2 = 9 x+2y+32 = 6 Sx+y+2 = 8 It can be written in matrix form as AX = B f2 3 1] fx] 9] wie as tae -|y,3-|8 laziest ls Let A=LU f1 0 0] a) eo vee 1 ca in fn 1) ye 83] 0 ug Us 0 tp ts 0 0 uss, 2021-37 (2) (8) fa) -D 42)98-2021 By equation 2) 32 0 Ojfun we os Faave Be? oe al ls 1 2 ird Semester, Computational Methods Thi " we get d bath 2, u 1 ‘Thus, L = | 3 3 2 2 ° ° ‘Then equation (2)--n(8)-00 +hylg = 1 D bys + lant Yn = 2 see 895 _,-3+35 og 8 3548495 ua alk 2 uw, = 18 0] [2 3 1] ofanau-|o 2 5 23 Bere loro 1s Ae Y, 1) 1} 3 lt vy al?| - x 1 |L?. ya} (2) becomes LUX = B Ly =B corresponding eigen vector of the matrix [ cig LP. University 1B Tech/-Akash Books 1ijg0 Oh ae A 1] fo Ya Te 5. 2 = y=9 a ii 9.3 pith = 6 = y,26-2-8 —"-Tyatys = 8 a7 2 = aT =e eno geeraaa 4 9 Xana 2 From (2), we get 23 eee 6 a ot Seema 1 | 2 oo ull] [5 = 2x+3y+2 = 9 35 = ee 18 = 1 2) oaemen = ee 18 182 = 5 5 eee 18 aD. 30 a6 1s" 18"" "18 Q.49. Explain power method. Determine the largest Eigen value and the 5 4 Jes: Ans. Power method: If X,, X, values i, Ayy soy then an arbitrary column vector ean be written as X = KX, +KX+.. + KX, AX = K,AX, + KAX,+... + KAN, ‘Then, ] using power method. 2021-39 , X, be the eigen vectors corresponding to the‘and the contribution of the therefore every time w, ves with rand < e ae to the eigen vector X, ‘Then we make becomes nearer Or nity to avoid the factor K,, gest root lution to possible 8 gives the firrt evaluate the solt which is a8 near tl mae after normalisation. as 1% aie eigen vector. Similarly We repeat this process ti and apremgtege valor and X, the corresponding becomes neg! ae {for finding the dominant eigen valve of a matrix is known as power ‘This process fo ‘method. Es Now, we determine the largest eigen vector ofl) 5 34 = he 1 Let the initial eigen vector be | _ fs 4]1)_[5]-5) 2 mS of pl (tc 0, the frst approximation tothe eigen value is 6 and the corresponding vector | (3 ox [f Sfoal-te)-*4oa = st[, 4] [oad -[i3]-#24] b= sseae,=[, 34] B 0.248 5 15.99) (i ale Bele 1°20 Ba B a, = 5.99andx,=| 1 0.248 LP. University-{B Tech}-Akash Books Beaty mo [T SJosu-[its)-*%f 02 Hence the largest eigenvalue is 5.99 and the corresponding eigen vector is [ a ne jus linear ber of ‘Ans. Guasé algebraic equation. This method produce the exact solution after a finite num! steps In this method, the unknowns are to an upper triangular system for which the unknowns are found by bac vely and the system is reduced i« substitution. For this method, produces is 1. Write down the equation in matrix form AX = B. 2, Write argument matrix 3. Convert the matrix A into an upper triangular form by using only row operations. Find unknowns by back substitution. mination, any one of the pivot elements ay, 8 --» My, vanishes or become very small compared to other elements in that row, then we interchange the { row (called the pivot now) with any other row below it, preferably with the row having greatest element in the #® column. is performed in (n ~ 1) steps in case of n unknowns. It is possible to count the total number of addition, subtraction, multiplications and divisions called ‘as operation count of the method. For Guass elimination method operation count for n system is n(n? + 3n ~ 1/3). r large n it is approx. n°/3, Q.51. Find the natural cubic spline to fit the data. =| me [ase ed yi | 10m mame [Sons ia) ‘Ans. Since the points are equispaced with h = 1 and n = 3 we obtain. Sept Mi. 4 4M, 4M) = 2.1 2y 49) 12 M,,#4M, +My = 69,.~29,+ Iq Jet = 12 M,+4M, +M, = 6(y,~2y, + 2y,) = 6-2) =—12 and M,+4M,+M, = 6%,-2y, +9) =6 Since M, = 0 and M, =0 = 4M, +M, = -1andM, + 4M,=6 On solving, we get 2 6 M, = Bou, -3: soon ‘Third Semester, vw M, = 300-5" 24 ie, ‘Thus the cubic splines T° 6 gbyax?-2x), «1x52 fix) = | x2-78x7+192-3, 26x53 4 cg? +12x 472460), 35x54 10 Q.52. Solve the system oxedyez= 3 Bx+2y-22 = -2 x-y+n= 6 by using Gauss Elimination method. Ans. mivigs2 = 8 Bx+2y-22 = -2 x-yte= 6 In Matrix Form: - t sow F g2 ‘ BTech)-Akash Books 20ni-43 artes x= 2y=-1en8 Q.53. Find the largest eigen value of the matrix, using P: ‘thod. ‘ower me [5 044-20 ser, Computational Methods : versity-B.T 5 0 yh 5.95 1 oe : Bees 0 Fl tal lee sy climinate 2m 9) and (10) pera 23 0 os os = are . = snhy ates I~ 550 nd [10)+ 2.02] 5 oD 1 5.95, a a ae e 1 sls le 2x, lowing set of equations using Crout’s method a " : saan e 21 Mt TX + 2x, + x, Bx, ox, + Ox, : = . ro negligible. sna 2 ilxele eile ia aes eae ce between 5.95 and 6.96 a! Since the differen FHanen the lrgos eigen value of muisix A 5.96 84, Solve the equation using Gav Bee agthod x39 +5515 = G at gy +z = 20, S-¥ + 82 = 10. ‘Ans. Given equations are ae ox -2y +52 = 13 ve aaete ax+Sy+4z = 20 pees gx-y+8z = 10 : aa Dividing (1) by 2, we get voles fe oes awa iy “ eee : e fan ta mo eliminate x rom (2) and (8) operate [(2)~ 214) and (8) ~ 341 12 ilies lee fav > wee 13 11 2 ree ne fattis * laatay Bava) 9: ©) ef : Latha + bya lay tys + battgy +h, by-2=7 eaaeen teen : 4 us due 4 oD 2 2 by = Paty Uy =e n= a es ly = Anky Uy = 1-1xd—a3 272 ) = 218)] eee 3 ines - ete 7 = seteus ere 7 ae Soar ao) agit Uae os 2 22°33 na aa) 20no) Dividing (1) by eee ” 70 So e oo a2) i eae é 22 Talontes 3 oo1 aE EGE oe erga er emaSemester Computational Methods ‘hird 4-202 £ ax=B,LU x=B,UX= w=-B 20 Oly) [7 sy 12 of] * [8 5 jee aa eae 3 ees yet 28 2 eS ae 2 Seno vysguerg’ mosey 8 35 Ss vel3 5 ux=V orl 2 2h.) [35 = 01 2fx} =| 8 fo o 1% * eee eS nttucjm = 95 1 ntim = 3 = 4-3 From eqn. (2) and (3), wehave x=2 = x,=1 x=1%22,%,=3 Q.56. Solve 6x + I5y + 552 = 76, 15x + 55y + 2252 = 295, 55x + 225y + 9792 = 1259 using Cholesky’s decomposition method. ‘Ans. We have 6x + loy + 652 = 76 Lox + 55y +2252 = 295 55x + 225y +9792 = 1259 Let AX =B [6 16 65) fs eee os) 241°], [55 225 979) [2] ] [76] B =| 295 | [1259 ( (3) ye Jay, = V6 = 2.4495 Now, LP. University-[B.Tech|-Akash Books 2021-47 225 - (24.4537 x6.1237) _ 99 9165 4.1833 bay = aga - Bi Ue Now, bing fn 0 1 be bs 24495 0 0 6.1237 4.18330 22.4537 20.9165 6.1101 (2.4495 0 oO LxL" = | 61237 4.1833 (0 22.4537 20.9165 6.1101, 2.4495 6.1237 22.4637 x| 0 4.1833 20.9165 ° o 6.1101 55 225 979. 6 15 55 = [15 55 225 Now, AX = B and A= LL" Lux = BLP. University-[B.Tech|-Akash Books 2021-49 rind she cubic splines and evaluate (1.5) and y(Q). 1n=3 “ = A yasty, + 41888%: = 25 een a asa, 0sTsss, EHO = PE By back substitution Frm) 2)=18 Eq (6) gives 6.1237(91.0268) + 4.18835, = 295 | = yy = 250988 Eq. @ gives 22.459731.0260) + 209165125 0998) + 6.1101y, = 125° = yy = 6101 Now, LX=Y ‘ 2.4495 6.1237 4.1583 | 0 = 24495x + 61 4 ease a.rssay + 7 @ enor = 61101 4 Again by back substitution Eq (7) gives g Pye Ba oat For =3in(,4I saves ae fix) = Zieae? + 24x? - 76x +80) = ye By (7, we have onde 2-4496x + 61297 + 22.4597 T+ 22.4597 = 31.0269 Also, y(@) = 44 trom both the splines of interval 2, 3 and [3, 41. xed ‘Thus, x= 1,y=1,2=1 Q.57. The following values of x and y areLP. University-(B ee eee 2021-51 s 03 = 14S [1+(0.3+1.4036)) = 1.4055 that y(0) = ay - 0giver 1 (0.2) ee (2014) = + [feo yan 1 9)+ fa ye ete” 2 Yo) im 9i°)] 3 2 Oe? “ae y = 47% a4y=4 2 1.4056)] = 1.4058 eae + (‘) = h series . HP = yor Zlflears0) + fe or’ a a som) ype a ys) = yor (2-%0)*" 26 = 1-23f1+003+1.4058) z a yor = 19844 as ye) = eee At x, = 0.3,y, = 1.4058 é a @.60 Use Runge-Kutta method 4th dy 2 144x0242x 002? of 4th order to solve $2 = 1+ 9%, (0) =01 ee : nd ¥(0.6) in 3 StePS as ys vl = 188 4 i & = = 19 «q.50 Apply modified Ealers method to Sad (0.3) give that ee mxeny Ans. Z y i) % = 0,y,=0,h= 02 ‘Ans. Let se =i0.3,519=03 for x, = 03 PEN fica Pee hy = hy Gy ¥_)=0.2%1= 02 1 Now Jothe flay ¥,)= 1408 = 18 me afeardined) 03418216 fd : A 0+0.2 YW,” = yotslfao¥0)*. (x, OP (0 : ) ary xo Yo) * f " : = 0.2£(0.1,0.1) = 0.2 (1 + 0.1%) = 0.202 4 +t) a, = Mtaehoo¥ 02 5, 0.202 oar (or50% 2 ) = 0.2f10.1, 0.101) = 0.2(1 + 0.10%) = 0.202LP. University-(B Tech|-Akash B = 0.2% 11785 = 0.2357 hy = riser fane ks) 2021-53 ozs (oa 92 04nan 92851) = 0.21(0.5, 0.54045) = 0.2584 h basic i{n+hone4) = 0.2105, 05516) = 0.2608 b= pete, 9) 2 0.2% 208 hy) hy = fl, +h, y+) h he? nflnry 3) = 0.2 §10.6, 0.6834) = 0.2904 02 » 9026+ 1 y +—— 0.2026 ig eee . « o2(00% a x, = 06 AL = 0A226 + 0.26125 = 06836 d Kuler’s method, find y(0.2) and y(0-A) corres 2 for x, y', = yet he Gy =1 4012, At = 0+ Us 30)* £3 14 921+ .2-0.2)) 1-216 2 = 1 Gn 0S i” * 02 11TH = 0.2367LP. University-(p, ‘Tech|~Akash Books 2021-65 Me™ Yet Yet MGry,)= 1+ 0,00), } 108 y (0.02) = 1.02 Sg, Y= ¥thlay) = 1.02+0.021 %tD 0.27) = 22 +c.2176-0 » = gape 2 - 104 = 1.0392 = 42177 y (0.04) = 1.00 392 emer h To ‘again Vy = ¥,thfayy,) Algood . 1oso2-0,02{ L0882=004) gp the 1.0392+0.04) 4 = 1.4524 = 3 0.9992 402 (1.217- 0.2") = 145 1.0392+0,02| ———— aa | leah 2 = 9 +Riflan 0 fGe” | = 10577 = 177+ (1.4624-0.4)) | y (0.06) = 1.0577 03 Apply the fourth order ILK method to solve Y teysyOul = | rs yO “ )+ flay PI) qeking h = 0.1 and determine approximation to y(0.1) and y(0.2) correct to 4 z ‘ decimal places. ay = 77+ (1.4634-0.4)] ‘Ans. Here fixy) = x+y',x,=0,y)=1,h=001 F 4 Now h, = hf yy.) =0.1(0' + 14) =0.1 pe It glflars0)+ Fa % ) . : 3 he {z.+hy,r8t)=03 (0.05,1.05) = ars 2a 177+ (1.46509-0.4)| greg yon = 1.4652 = 0.11(0.05)? +(1.05)?] = 0.1105 ‘Thus at es fe a x, = 049, a is w(erhoes Q.62 Solve the differential equation ee by Euler’s method with the imitial . = a 7] = 0.1116 condition y(o) = 1 for x = 0.06 taking interval of differencing h = 0.02 0.1(0.05)* + (1.0553) 1 1116 (2018) hy = hf la, + hy y+) Su ® - puz,y) 2% = 0.1£(0.1, 1.1116) = 0-1 (0.12 + (1.1116) es ae } = 0.1246 *, = Oy,=Lh=0.02LP. University{B ‘Tech}-Akash Books i wai-37 s Methods Y = %+ Ife, ide © 1s fix-tdewt. 2, 1246) ; n yams To find Lue 01 " s a\* I | i Si ah oat eoiaal : Q.65 Use Das or's series method to solve the equation & =-sy 910) at ee Oh Ans. Let fx,y : . ssa Here that yl Q.S1 Use Picard method w approximate y when » = 02 saiicaeat ®t ven © . 1-7-9 Ane 2 j | — Pao a By Pears method sLP. University-{B Tech]-Akash Books 2021-59 bs ofaehovel = 0.1F(0.15, 1.1598) = 0.1144 Peake a hy = Atl +4, y+% flat genta) = 0.10.15, 1.1621) = 0.1145 A, = hflx +h,y, +h) = 0.1 (0.2, 1.2194) = 0.1191 Ascot on ort 02... 1 pies 62 k= 2h, +2h, +2, +h,) x 2 5 it Bh + 2h, + hy = 0.9948 (ApPrOX)- a Seder order msetoed t9 S0t Yun TUNES = 2 (010.6 + 20.1144 + 2% 0.146 + 0.1190 Kul Co es ay erg YOr* 2018) = 0.1144 : roblem » 5 = for the initial value P? ae At x, = 02, y,=11049+ 0.1144 dy 2 gry =f) = 1.2193, aie ate oy,2 b=? ie heey) Q.67 Using fourth order R - K method to solve 2 = ay+y*,x(0) = 1, Take ae fern OMOFE = 04 interval h = 0.1. Compute y(0.1) and y(0.2). (2017) =A ee ‘Ans. Here ftx, y) = xy + y* h=0.1,x,=0,y,=1 h)-oar(S 1492) : : ow, 2 h wu(s0+5190* 7 1.05) = 0.1 x 1.0488 = 0.1049 > " )=0.10+1)=0.1 = 0.1 (0.05, ) K, = Re eee hy ueorF nr) ¥ of aie ze ae oe: 3. : ons ca a ne = 0.1 10.65, 1.0525) = 0.1 x 1.05 oap(or 92.1 +22) 0:1710.05.1.06) = 0.105 = 0.1(0.05 x 1.05 + 1.052) = 0.1155 [eee hy = hile, +h, p+ #) K, = ht (xy + "904 == 2 2 0.10.1, 1.105) = 0.1098 01 aflo+ 21+ oap{o- 2a oa 2 b= 1 (y+ 2h,+ 2h, +h) 6 = 0.1f (0.05, 1.0578) = 0.1172 = $ (0.1 + 0.10492 + 2 «0.105 + K, = hffx,+h,y,+K,)=0.1 10.1, 1.1172) = 0.1049 ee At x, =O, y,= 140.1049 = 1.1049 K= 2K, + 2K, + 2K, + K) At m= 02 = 1 (01+2%0,1155 + 2 x 0.1172 + 0.1360) hy = MG, 1 fO.1+ 1.1049 | § = 0.10982021-61 (0 + log (0.2 + 2.0656)] | = 21415 = Loss9+2*0 1582 + 2x 0.1610 ~ 0.1889 K = 0.1605 = x, = 02 3 y,ey, 7K = 11169 + 0.1605 i ye . at ee y2= 21416 ‘Ans. Let x, =0,¥,=2,b= Thus x 02 o4 for x, y 2.0656 2.1416 (x+y) es fay get ae Q.69 Apply Range-Kutta method of fourth order, solve 77 = = with = log (0 +2) = 0.3010 Pox Now se yO) =Latx=02, 04. eo) “% =2+0.2x 0.3010 = 2.0602 (2.2602) = 0.3541 Ans. fix,y) = 2%, x) = 07 = Lh = 02 yee “ 1-0 = yee foyy) = 597) - 10-8010 + 0.3541) = 2.0655 &‘Third Semester, Computational Methods ky, = bf (y yo) = 0.2 x 1 = 0.2 LP. University ky o2rlog 2021-09 kee (sabre +B) = 02F0+ 0.1403, (0 a? - 0.17) = I) = 0.25 0.2/0.1.) = 020753, = 0.1967 k, x hy. 2 ( 0. bi k, = i(x+4.00+4 o2f ek sae s (1.0984? ~ 0.12) 2 = 1,1.0984) = 0.2) =POS4 — 0.1" pai (1098+ 0.2) | x = 0.1967 k= k, = hf, +h,y, +k) =0.200+0.2,140,: i 0.1891 + 20.1795 + 20.1705 + 0.1686) 1.1967? — 0,22 = 0.2/(0.2,1.1967) = 0.2| 21967 = 0.2") latiser? +07) = 0.1793, At x, = 0.4, y, = 1.196 + 0.1793 = 1.3753 = 0.1891 Q.70. Using Range-Kutta ethod of Order 4, find y for x= 0.1, 0.2, 0.3. Given that: + hy + Dy + hey) alm ie = Ho2+2x03867 + 20.1067 + 01400 Continue the station sf 01 eaae Maleate = 0.1960 Atx, =0.2,y,=y, +k = 1+ 0.1960 = 1.196 E ¥) = (0.10.1) = 0.1 Now x, = 0.4, y, = 1.196, x, = 0.2, h = 0.2 1 1 Jn yp +k aoa 1.1967 + 0.2 k, = hf (x,,y,)= 0.2 x 0.9456 = 0.1891 = (0.1) £11.05, 1.05) = 0.1185 an ‘oo 14) (era 0.1891 Si 10577 = 0.172 = Milas doo +4) -o2/(02+ 01,1 196 + vce heiga ky <2) 051, LID = 02388 | [1.2206 ke = Ne, + ey +25 +s) = 0.2/(0.3,1.2906) 02| See | Os ae: 1 “0.9949 + 0-19598)= 0.11687 . = 0.1795 = Lox +02m y,= yor k= L169Computational Methods ‘hird Semester 4 64-202) ro find y(0-2) ee Dee ee ey es) 0.1, 1260) = 0.1800 arenes een ese es) ne age 2) = (0.100. 15, 1.1848) = 0.1681 1 ives) a wfaedane ds) (0.1) 0.15, 11959) = 0.1609 bit, +9, *15))= (0) 02, 1.2778) = 0.1055 ee De, + Bky + 2k +h) =O- 1605 é Thus oa) = ¥,29¢k= 12738 Tosfind (0.3) Here x, = 0.2, y, = 1.2778, h = 0.1 k= = (0.1) 80.2, 1.2773) = 0.1887 ee neta = (0.1) 0.25, 1.8885) = 0.2275 k, = hftx, +h, y, + k,) = (0.1) 0.3, 1.5048) - 0.2716 k= Ue tak, 42k; +k,)=0 Thus 03) = y,=y,+k= 1504 Now the starting values for the Milne’s method are: oe p= 1.0000 oa = 11169 ep Y= 12773 ee ¥,= 1.6049 Using the predictor 4h 0° @h -f, +21) LP. University-[B.Tech|-Akash Books 2021-65 = 12773+ 211.8869 + 4(2.7132) + 4.098) f= in vsing the corrector . ae « [1.8869 + 4(2.7132) + 4.1159] = 1.8391 = 4.1182 musing the corrector : Ys y0.4) Hence Solve the elliptic 7. cataith boundary, values or shown: im Ans. 500 1000 500 0 lc 0 5 Up | Us |amaOy 1000) 1000 A 8 Uy ean 2000 2000 Up | aU |e 1000 2 1000 ° o 500 1000 500 ios Let u,, U,, U, ..., u, be the value of u at the interior mesh-points. Since the boundary values 1 are symmetrical about AB u, = U,u,=u,u, =u, Also, the value of u beign symmetrical abou: CI ‘hus it is sufficient to find the values u,, U,, Uy U, 6000 = 211000 + 2000 + 1000 + 2000} == =1500 (0 +1500 + 1000 + 2000}= 714500] = 1125 {1000 +1125 +1500 +1125] = 1188Methods Computational sthird Semester 1g - 2000+ 1125+ 15001= 1438 LP. University-[B.Tech|~Akash Books 1 oes p aval interior mesh PIS. We Will imp values 4 = Lt1000-+ 1063-+ 500-+ 1818) = 969 Now, we have got the Toush £5 ee method (or SFPF). She values by using Gauss-seldal ae = i +1000 +1201] = 1038 First Iteration: : aie) cn = 24500+1000+ uf +2 u,® = 112000+1201+ 969 +982] = 1288 any 4 = 24500 + 1000+ 1438+1188) = 1032 aye = 21n268 -1918- 1038 +1063) -1176 4 4 uO , eo: ues Stroo0 +o +99 sul) Fifth Iteration: a u,° = +11000+ 1038 + 500 +1288) = 957 14,000 + 1082+ 1500+ 1125] = 1164 A a 8 + 500+1: 9 4 1 es, 4, = 11957 ~-969+ 1000+ 1176) = 1026 yo = ful? + 2000+ uy” +05") 4 MeL eel a v= 212000 +1176 +957 + 969] =1276 = Lszoge +2000 +1125 + 1500)=1414 4 4 1 : u,® = 2i1276 +1288 + 1026+ 1038] = 1157 are euP sul +00) 4 eee larly u,© = 951, u,= 1016, u,®= 1266, u,® = 1146 = Apes --1414~ 1188 ~ 1438] = 1301 uy? 4 ae Second Iteration: 4 u.® = 1000 +116: +500+1414]= 1020 4” used 4 2) = 1128 a? = 2,y020 = 10521000 -13011= 2088 wu uj! = 1001, u,"? = 1251, u,® = 1126 Thus there is negligible difference between the values obtained in the 11 and 12 1 = 212000 + 1301+ 1020 + 1032] = 1388 iterations, 4 = 1001, u, = 1251, u, = 1126 ee Arnage + 1414 +1088 +1164) = 1251 Q.72. Apply Adams-Bashforth Method, to find a solution to the differential a - ‘Third Iteration: | equation gexta tx = 1.4 given y(1) = 1, y(1.1) = 1.233, y(1.2) = 1.548 and | y3)= 1.979. u,® = 1x00 + 1088 +500 +1338] = 982 ae aot Ans. Here h = 0.1 and fix, y)= x41 + y) ee . %=Lyo=1 u,? = 11962+ 1020 +1000 + 1251] = 1063 4 fixyy)) = x+y) =2 a.1 Sy ans 8 = 712000 +1251 + 982+ 1020] =1313 nL) ice fix, y,) = x1 +y,)= 2.702 pee 9 = 711313 +1338 +1063 +1088] = 1201 Ba 2s Yaceoae 4 Fourth Iteration: fixyy) = xf +y,)=3.669 — ll eanal Methods 1) Use predictor bh ye 37fley 91)~ 5942-72) * pe yy e Peloton V0 y 02987 gat 0, 02-5: é DA gna 37427 ype 1979+ 94! 9 x 3.669 + 55 + 9flx, Use corrector formula = 7.004 702-5 3.669 +19 are all zero. Final value at x= 1418 ues of the following 73. Evaluate the pivotal val to one half of the period of vil and up ps0 > Us dge = 0 OF Ua rations 1 given u(0, 0) = 05, u(x, 0) = x45 -») a ‘Ans. 2 _ g2 20 is the equation of vibra Hey uation of vib vibration = 2 a Here! = 5, 0° 16(u,,~2u, + u YF we got Taking h = T6k%u,. — 20, + @ Uy) = 21-16k)u,, + 16k, Choose ‘k, 80 that coefficient of u, = 0Computational Methods 70-2021 ‘Third Semester, Puti=1,2,3,4in 6), we get Uy = Var t Yor a2 Peon k eA ne = U4, +%,1— M20 PS eee as sus r+ ods 20 uy + U2,17 Yao Pesaran 2 = 18)= 10) = UW tY, Ugo = Usrt Yara Meo =0+18-16=2 = Uy ot Ys,07 Yao The table is as follows: | =2 -10 -10 ai elololelololo | =16 -18 -12 -4