Research Journal of Recent Sciences _________________________________________________ ISSN 2277-2502 Vol. 4(1), 73-80, January (2015) Res.J.Recent Sci. International Science Congress Association 73 Homotopy Type Methods for Numerical Solution of Non Linear Riccati Equation Naeem M, Badshah N, Shah I A and Atta HDepartment of MathematicsIslamia College Peshawar University, PAKISTAN Department of Basics Sciences University of Engineering and Technology Peshawar PAKISTAN Department of MathematicsIslamia College Peshawar University PAKISTAN Department of MathematicsIslamia College Peshawar University PAKISTAN Available online at: www.isca.in,www.isca.me Received 4th December 2013, revised 2nd March 2014, accepted 30 November 2014Abstract In this paper, we apply various homotopy type methods for the approximate solution of nonlinear Riccati differential equations, such as Optimal Homotopy Asymptotic Method (OHAM), Homotopy Perturbation Method (HPM) andHomotopy Analysis Method (HAM). We also comparethe results of each of them with the exact solution of the given problem. Conclusions reveal that the method OHAM is more effective and the results obtained by this method are in good agreement with the exact solution. Keywords: Nonlinear Riccati differential equations; OHAM; HPM; HAM. Introduction A nonlinear differential equation which is useful in the dynamics of a system is given below and which is called Riccati equation 2 ()()()0 dvazbzvczv dz ---= (1) Where () vz is unknown function and ()0 az ¹ , ()0 cz ¹ . This equation is named after the Italian mathematician, Riccati (1676-1754), the recent applications of the given equation involve such type of areas as financial mathematics1,2. The solution of this equation is found using the classical numerical methods such as the forward Euler and Runge-Kutta techniques. A scheme which was presented by Saidiet al was stable. The Riccati equation solution is independent of the solution obtained by the techniques that produce solutions to linear differential equations. The analysis is less efficient as well. Moreover because of similarity of the Riccati equation to the equation analyzed in chaos theory, there is further investigation that involves the concept of chaos theory but there exists simple analysis of Riccati equation. If one solution of Riccati equation is known then a whole family of solutions can be found. For solving this equation no analytical method exists. A method for solving this equation numerically is discretization of it in time domain. Such type of equations can be solved by other methods e.g. Adomian Decomposition Method4--7 and Homotopy Perturbation Method8-11, but we use OHAM, HPM and HAM for the solution of Riccati nonlinear differential equations, results are compared with their exact solutions. The OHAM results are reliable and show excellent agreement with the exact solutions. Basic Idea of Oham: We consider the following differential equation. (())()(())0,,0 dvSvzfzEvzDvdz ++==  ( ) 2 Where S is a linear operator, () vz is an unknown function. () fz is a known function, E is a nonlinear operator and D is a boundary operator. According to OHAM, a homotopy is constructed as ((,),):[0,1] YzppRR q ´® that satisfies ()() ( ) ()()()()()() 1[((,)) (),], ,,0SzpphpSzpfzfzEzpzpDzp-= \n \n ++ (3) Where zR Î and [ ] 0,1 pÎ, () hp is a non zero auxiliary function for 0, h(0)0 p ¹= and ( ) , zp is an unknown function. Obviously, when 0 p = and 1 p = it holds that ( ) ( ) ,0 zvz and ( ) ( ) ,1 zvz . As p varies from 0 to 1 , the solution ( ) , zp approaches from ( ) 0 vz to () vz where ( ) 0 vz is obtained from equation (3) for 0 p = and we have ()()() 0 0,,0 dvSvzfzDvdz+== (4) Research Journal of Recent Sciences _____________________________________________________________ ISSN 2277-2502Vol. 4(1), 73-80, January (2015) Res.J.Recent Sci. International Science Congress Association 74 ( ) 5 Where 12 ,, CC  are constants to be determined. By using least square method. () hp can be expressed in many forms as suggested by V. Marinca et al12--16. For getting an approximate solution, we expand ( ) ,, i zpC in Taylor’s series about p in the following pattern ()()()012,,,,,, k ikk zpCvzvzCCCp =+(6) Using equation (6) into equation (3) and equating the coefficient of like powers of p we obtain the following linear equations. Zeroth order problem is given by equation (4) and the first order problem is given by equation (7) 1100 (())()(()), ,0 LvzgzCNvz dvBvdz +=  ( ) 7 The general equations for () k vz are given by 00011(())(())(())(()) ((),(),....,()) 2,3,...,,0kkkikikikLvzLvzLvzCNvzC Nvzvzvz dvkBvdz-- -=  + \n + ==(8)Where 01 ((),(),...,()) mm Nvzvzvz is the coefficient of m p in the expansion of ( ) ( ) , Nzp about the embedding parameter p . ( ) ( ) ()()()0001,,,,mmNzpCNvzNvvvp ( ) 9 It suggests that the convergence of the series (6) depends upon the constants12 ,, CC  If it is convergent at 1 p = , one has()()()01,,,,ikk zCvzvzCC =+ ( ) 10 The result of themth order approximations are given ( ) ()()12012,,,,,,,,iivzCCCvzvzCCC   (11) Using equation (11) into equation (2), we have the following residual: ( ) ()()()()()121212,,, ,,,,,,mXzCCCSvxCCCfzEvzCCC++   ( ) 12 If 0 X = is taken, then v  will be the exact solution. In order to find the values of i C , 1,2,3, i =  We first construct the result ()()1212,,,,,m TCCCXzCCCdz  ( ) 13 and then minimizing it, we have 12 0,0,,0 TTTCCC ¶¶¶ === ¶¶¶ ( ) 14 With these constants known, the approximate solution (of order m ) is found.Numerical Problems: Problem 1: Considering the following first order differential equation on domain0,1 2 ()10,(0)0 dvvzv dz -+== ( ) 15 The exact solution of the problem is () vzTanhz =- ( ) 16 Applying the technique,OHAM theZeroth order problem is 00 ()10,(0)0 vzv +== ( ) 17 0 () vzz =- ( ) 18 First order problem is (19) 3 111 (,) 3 vzCzC - (20)Second order problem is 2110111 (,)2()()(1)(), (0)0 vzCCvzvzCvz v ¢ =-++ (21) ()3325221111(,)552 15 vzCzCzCzC=--- (22)Third order problem is ( ) ( ) ( ) ( ) ()'2'311110212 (,)21, 00 vzCCvCvzvzCvz =--++ ( ) 23 332521113133537311110521084(,)3151058417zCzCzCvzCzCzCzC  ---   ---  ( ) 24 Fourth order problem is (,)2()()2()() 41112103(1)(),(0)0 134 vzCCvzvzCvzvz Cvzv=--++= (25) 4133252111335373341111547494111(,)94528351134283522684599452835113445962vzCzCzCzCzCzCzCzCzCzCzC =  ---  ----   ---  (26) ( ) 2 12 hppCpC=++  1111010 (,)1()(1)(), (0)0 vzCCCvzCvz v ¢ =+-++ Research Journal of Recent Sciences _____________________________________________________________ ISSN 2277-2502Vol. 4(1), 73-80, January (2015) Res.J.Recent Sci. International Science Congress Association 75 Fifth order problem is (,)2()()112113 2()()(1)(),(0)0 10414vzCCvCvzvzCvzvzCvzv=-- -++= (27) 51332521113353731113454741119435551117595115111(,)51975207900831603118502494805049020790024948010098015592513640519758316050490136401382vzCzCzCzCzCzCzCzCzCzCzCzCzCzCzCzC =  ---  ---   ---------      (28) Using equations (18), (20), (22), (24), (26) and (28) the fifth order approximate solution for 1 pis = 101121 314151 (,)()(,)(,) (,)(,)(,) vzCvzvzCvzC vzCvzCvzC =++ ++  (29) Following the technique, OHAM l on the domain [0,1] we get 1 Xvv =-+  (30) The value of 0.7736625644632285 C =- is obtained, and ourapproximate solution is 357 ()0.3331350.1319010.0496428 911120.01492860.00245669() vzzzzz zzOz=-+-+++ (31)The following table-1 exhibits values of the variablez exact solution, OHAM solution, HPM Solution, HAM solution and their absolute errors. Also we investigate that there exists relation between their values. From the figure-1 we conclude that the two graphs1 (a) OHAM and exact1 (b) HPM and exact are coincident but in 1(c) HAM and exact are not coincident. Red curves of each graph show exact solutions while the green curves of each graph exhibit series solutions however, in general the method OHAM is more effective than the others. (a) OHAM and exact solutions graph (b) HPM and exact solutions graph(c) HAM and exact solutions graph Figure-1 The variable Z is represented along the horizontal line and the function v(z) is represented along the vertical lineTable-1 Values of the variablez exact solution, OHAM solution, HPM Solution, HAM solution and their absolute errors Variable z Exact Solution OHAM Solution HPM Solution HAM Solution Error OHAM Error HPM Error HAM 0 0.0000 0.0000 0.0000 0.0000 0 0.010 ´ 0 0.010 ´ 0 0.010 ´ 0.1 -0.099668 - 0.0996682 - 0.099668 -0.0996882 7 1.810 - ´ 16 3.610 - ´ 7 1.810 - ´ 0.2 -0.197375 - 0.197376 - 0.197375 -0.197376 6 1.210 - ´ 12 2.910 - ´ 6 1.210 - ´ 0.3 -0.291313 - 0.291315 - 0.291313 -0.291315 6 2.710 - ´ 10 5.510 - ´ 6 2.710 - ´ 0.4 -0.379979 - 0.379952 - 0.379949 -0.379952 6 3.510 - ´ 8 2.310 - ´ 6 3.510 - ´ 0.5 -0.462117 - 0.46212 - 0.462117 -0.46212 6 2.910 - ´ 7 4.010 - ´ 6 2.910 - ´ 0.6 -0.53705 - 0.537051 - 0.537045 -0.537051 6 1.610 - ´ 6 4.110 - ´ 6 1.610 - ´ 0.7 -0.604368 - 0.604369 - 0.604339 -0.604369 7 8.710 - ´ 5 2.910 - ´ 7 8.710 - ´ 0.8 -0.664037 - 0.664038 -0 .66388 -0.664038 7 9.210 - ´ 4 1.510 - ´ 7 9.210 - ´ 0.9 -0.716298 - 0.716299 - 0.71561 -0.716299 6 1.110 - ´ 4 6.810 - ´ 6 1.110 - ´ 1.0 -0.761594 - 0.761594 - 0.759038 -0.761594 7 1.810 - ´ 7 2.610 - ´ 7 1.810 - ´    0.2 0.4 0.6 0.8 1.0 0.7 0.6 0.5 0.4 0.3 0.2 0.1  EXACT 0 00000000000000000000000000000000 0 0.2 0.4 0.6 0.8 1.0 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 OHAM    0.2 0.4 0.6 0.8 1.0 0.7 0.6 0.5 0.4 0.3 0.2 0.1  EXACT 0 00000000000000000000000000000000 0 0.2 0.4 0.6 0.8 1.0 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 HPM    0.2 0.4 0.6 0.8 1.0 0.7 0.6 0.5 0.4 0.3 0.2 0.1  EXACT 0 00000000000000000000000000000000 0 0.2 0.4 0.6 0.8 1.0 1.0 0.8 0.6 0.4 0.2 0 HAM Research Journal of Recent Sciences _____________________________________________________________ ISSN 2277-2502Vol. 4(1), 73-80, January (2015) Res.J.Recent Sci. International Science Congress Association 76 Problem 2: Considering the following nonlinear differential equation on domain [0,1] 2 1030,(0)0 dvvvv dz --+== (32) The exact solution is 14()2 52 z z vz e =-+ + (33) We apply the technique, OHAM Zeroth order is 00 ()10,(0)0 vzv ¢ == (34) ()10 vzz = (35) First order problem is 1111010101 (,)10103() ()(1)(),(0)0 vzCCCvz CvzCvzv ¢ =--- +++= (36) ()2311 (,)920 3 C vzCzz =-+ (37) Second order problem is 2111101112 (,)3()2()() (1)(),(0)0 vzCCvzCvzvz Cvzv ¢ =-+++=(38) 232122345[920](,) [9296080] CzzvzC Czzzz  -+   +-+-+  (39) Third order problem is (,)()3()311112 2()()(1)(),(0)0 102123vzCCvzCvzCvzvzCvzv=- +++= (40) 23233145234567756168015124872(,)1008013440252756319210647217562856027200zzzzvzCzzzzzzzz   -+   -+  \n  -+   -+  + \n  +-+  (41) Fourth order problem is (,)2()()3()411213 2()()(1)(),(0)0 103134vzCvzvzCvzCvzvzCvzv=- +++= (42) 41232345232456734356(,)3402075600[102060328860680400907200]10206043092014373452937060385560036720002268177660756945204516942978606588000669vzCzzCzzzzzzCzzzzzzCzz = -++-+-+-+ \n+-+ \n \n-+ ++-+-891060004960000()zzOz              \n  \n  \n +    +  (43) Fifth order problem is 511211314104145 (,)()2()()3() 2()()(1)(),(0)0 vzCCvzCvzvzCvz CvzvzCvzv=+-+++=(44) 2322345323456742351[7484401663200][299376096465601995840026611200][44906401896048063243180129230640 169646400161568000] [299376015634080(,)666111601749896CzzCzzzzCzzzzzzCzzvzCz-++-+-++-+-+-++-+=-+678952345678910119974872378211680579744000 589248000436480000] [74844046569602332638077355432208733679427317660691891200860992000729696000442240000]zzzzCzzzzzzzzzz -++ -+-+-+ -+- +                           (45) Utilizing equations (35), (37), (39), (41), (43) and (45) we find fifth order approximate series solution for 1 p = is 101121 314151 (,)()(,)(,) (,)(,)(,) vzCvzvzCvzC vzCvzCvzC =++ +++  (46) Following the technique, OHAM on domain [ ] 0,1 we get the residual 2 103 Xvvv ¢ =--+  (47) The value of 0.3641329904497182 C =- is obtained. The approximate solution is 234 5678 9101112()1013.440720.892256.9436 37.7184133.79760.0512 170.756 81.77593.621756.7404()vzzzzz zzxz zzzOz=+--++--++-+  (48) The following table 2 displays the values of the variable z , exact, OHAM, HPM and HAM solutions and errors. From the table 2 we conclude the absolute errors of OHAM and HAM are approximately the same while the errors of HPM are larger than the errors of the other methods i.e. OHAM and HAM. From figure 2 (a) we admit that the graphs of exact and OHAM solutions are coincident while in 2 (b) and 2(c) the graphs of exact and HPM, exact and HAM solutions are not coincident but generally we investigate that the method OHAM is more effective than HPM and HAM Research Journal of Recent Sciences _____________________________________________________________ ISSN 2277-2502Vol. 4(1), 73-80, January (2015) Res.J.Recent Sci. International Science Congress Association 77 (a) Graph of exact and OHAM solutions (b) Graph of exact and HPM solutions (c) Graph of exact and HAM solutions Figure-2 Problem 3: Considering the following nonlinear differential equation on domain 01 x ££ 22 4470, (0)0dvzzvvdz -+-+= (49) The exact solution is ()23 z z e vzz e  =+  +  (50) Applying the OHAM technique, mentioned above, Zeroth order problem is 00 ()47,(0)0 vzzv =-= (51) 3 01 ()(214) 3 vzzz =-+ (52) First order problem is 22 1111 2 101010 (,)7474 4(1), (0)0 vzCzCzC zCvCvCv =-+- ¢ +-++ (53) 57 11 8085(,)315151280zCvzC zCzC  -=   +-  (54) Second order problem is 211110111 2 (,)42(1), (0)0 vzCzCvCvvCv v ¢¢ =-++ (55) Table-2 Values of the variable z , exact, OHAM, HPM and HAM solutions and errors Variable z Exact Solution OHAM Solution HPM Solution HAM Solution Error OHAM Error HPM Error HAM 0 0.0000 0.0000 0.0000 0.0000 0 0.010 ´ 0 0.010 ´ 0 0.010 ´ 0.1 1.12296 1.10832 1.12296 1.10832 6 1.510 - ´ 8 3.110 - ´ 6 1.410 - ´ 0.2 2.33036 2.29886 2.33035 2.29886 6 3.210 - ´ 6 9.110 - ´ 6 3.110 - ´ 0.3 3.3593 3.35125 3.35957 3.35125 7 8.010 - ´ 8 2.710 - ´ 7 8.010 - ´ 0.4 4.07626 4.10851 4.07582 4.10851 6 3.210 - ´ 8 4.310 - ´ 6 3.210 - ´ 0.5 4.50864 4.54624 4.50098 4.54624 6 3.710 - ´ 7 7.710 - ´ 6 3.710 - ´ 0.6 4.74706 4.75674 4.78838 4.75674 7 9.710 - ´ 6 4.110 - ´ 7 9.710 - ´ 0.7 4.87207 4.86157 4.20166 4.86157 6 1.010 - ´ 5 6.710 - ´ 6 1.010 - ´ 0.8 4.93588 4.9274 -9.61142 4.9274 7 8.510 - ´ 0 14.510 - ´ 7 8.510 - ´ 0.9 4.96801 4.9659 -113.087 4.9659 7 2.110 - ´ 0 118.110 - ´ 7 2.110 - ´ 1.0 4.98408 4.96981 -624.264 4.96981 6 1.410 - ´ 0 629.210 - ´ 6 1.410 - ´            0.2 0.4 0.6 0.8 1.0 1 2 3 4 5  EXACT 0 00 0 0 0 0 0 0 0 0 0 0 0.2 0.4 0.6 0.8 1.0 1 2 3 4 5 0 OHAM            0.2 0.4 0.6 0.8 1.0 1 2 3 4 5  EXACT 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0000000 0 0.2 0.4 0.6 0.8 1.0 1000 800 600 400 200 0 HPM            0.2 0.4 0.6 0.8 1.0 1 2 3 4 5  EXACT 0 00000000000000000000000000000000 0 0.2 0.4 0.6 0.8 1.0 2 4 6 8 10 0 HAM Research Journal of Recent Sciences _____________________________________________________________ ISSN 2277-2502Vol. 4(1), 73-80, January (2015) Res.J.Recent Sci. International Science Congress Association 78 35117321152722111921121113266805498962640266805(,)9106022273041039520064640()zCzCzCzCvzCzCzCzCzCOz  -+  --   =-+   -+   +  (56) Third order problem is 311112102123(,)4 2(1),(0)0 vzCCvzCvCvvCvv=-+ -++= (57) (58)Fourth order problem is 4111213 103134(,)24 2(1),(0)0 vzCCvvzCC CvvCvv=-+ -++= (59) 412711(,)17644948571253299834538001745944200052934845713751806659409555045097738686003980752776012697776000529348457137568746552875vzCCzCz \n \n \n  - \n - \n \n ++ \n \n - \n + \n 1113157123138552500593400602544751962092091960024083871739201379173824003144211200176449485712518726561003150638672935650756181995574755026884144  \n \n \n - \n \n + \n - \n \n + \n \n - ++1113151719024920427655235168034714969190414675904000260249600                              \n  \n  \n  \n  \n  \n  \n  \n - \n  \n +  \n  - \n  \n  +  (60) Fifth order problem is 213511145404 (,)(1),(0)0 42vvvvzCCCvvzvvv--¢¢ =++= \n+- (61) Table-3 OHAM solution, HPM solution, HAM solution and their absolute errors z Exact sol OHAM sol HPM sol HAM sol Error OHAM Error HPM Erro r HAM 0.0 0.000000 0.000000 0.000000 0.000000 0 0.010 ´ 0 0.010 ´ 0 0.010 ´ 0.1 -0.673938 0.677862 -0.673938 0.677862- 7 3.910 - ´ 9 1.510 - ´ 3 3.910 - ´ 0.2 -1.21115 -1.23345 -1.21114 -1.23345 6 2.210 - ´ 5 1.110 - ´ 2 2.210 - ´ 0.3 -1.54889 -1.59119 -1.54712 -1.59119 6 4.210 - ´ 3 1.810 - ´ 2 4.210 - ´ 0.4 -1.70096 -1.74404 -1.64022 -1.74404 6 4.310 - ´ 2 6.110 - ´ 2 4.310 - ´ 0.5 -1.71544 -1.74015 -0.834453 -1.74015 6 2.410 - ´ 1 8.810 - ´ 2 2.410 - ´ 0.6 -1.64042 -1.64509 5.79423 -1.64509 7 4.410 - ´ 0 7.410 - ´ 3 4.710 - ´ 0.7 -1.51136 -1.50596 41.7794 -1.50596 7 5.310 - ´ 0 43.210 - ´ 3 5.410 - ´ 0.8 -1.35102 -1.34184 191.103 -1.34184 7 9.110 - ´ 0 192.410 - ´ 3 9.210 - ´ 0.9 -1.17302 -1.16033 695.585 -1.16033 6 1.210 - ´ 0 696.710 - ´ 2 1.210 - ´ 1.0 -0.985164 -0.956882 2144.38 -0.956882 6 2.810 - ´ 0 2145.310 - ´ 2 2.810 - ´ 3135117321152721192112391111(,)364188825681080403603600728377650248594346062053992054774720364188825255405150014189175408256348517472001349908560165695712vzCzCzCzCzCzCzCzCzCCz-+---+++131516488640216320()Oz          \n  \n  \n  \n  \n  \n  \n  \n  \n  \n     Research Journal of Recent Sciences _____________________________________________________________ ISSN 2277-2502Vol. 4(1), 73-80, January (2015) Res.J.Recent Sci. International Science Congress Association 79 (,)9374761175905125175320209003940092762015346000374990447036205001279837525728762003194723808516240028199652665184001189951045184000365250435424875vzC\n\n\n\n\n 5624856705543075039474047025886500063054748026405135020849190569166960011255915221100739201314655101053824001533410388277200  \n \n \n \n \n \n \n \n \n  - \n \n - \n \n - \n \n ++ \n \n - \n \n + \n - \n \n 37499044703620500397976874438943800 7 1357307722844973900131379769954692600 11 57134182881759984013908852905779033601573776252523438081173118923118080001955308244992009374761175905125133243358842994400694720242480414150 9 1525109570505120600 11 963379587114314370 13 559528566702724080 15 1110089128698728541711668837527857664707  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n 1921102053324821236737040179202334348421120025()Oz                      \n  \n  \n  \n  \n  \n  \n  \n  \n  \n  \n  \n  \n  \n  \n  \n  \n + \n  \n - \n  \n  +                                                              (62) Using equations (52), (54), (56),(58), (60) and (62) we get the following approximate solution 101121 314151 (,)()(,)(,) (,)(,)(,) vzCvzvzCvzC vzCvzCvzC =++ +++  (63) Following the technique, OHAM on domain [ ] 0,1 we use the residual. is more effective than HPM and HAM. 22 744 Xvzzvv =+-+-  (64) The value of C is -0.29729358104456594 and the approximate solution is () 5357911738587792403.6211214449.508331523.783555738.395315.920874.81773.331039556.5654534.638101.72855101.6319104.0311210141891754.5082210vzzzzzzzzzzz=-++-++-´-´+´-´+´ 1113151611311512712911111013915817619249322.05683.99()1.82024108.65734101.46408101.03365102.74414103.7030810687465528752.79442101.14643102.0329810(zzzOzzzzzzzzzzOz++´-´+´-´++´-´+´-´+´+ 211431551671691511151314151317121910216.79581104.26724101.04299101.24486107.38081102.044789103652504354248753.16311102.95355101.6856105.49786107.9769zzzzzzzzzz´-´+´-´+´-´+´-´+´-´82325110()zOz´+           (65)Table-3 displays the values of exact solution, OHAM solution, HPM solution, HAM solution and their absolute errors, the values due to OHAM and HAM are nearer to exact solution but HPM values do not agree with the exact values. From the figure 3(a) we conclude that the exact and OHAM solution graphs are coincident exhibiting that OHAM, technique Research Journal of Recent Sciences _____________________________________________________________ ISSN 2277-2502Vol. 4(1), 73-80, January (2015) Res.J.Recent Sci. International Science Congress Association 80 (a) OHAM and exact solutions graph (b) HPM and exact solutions graph (c) HAM and exact solutions graphFigure-3 The variable and  are represented along the horizontal and the vertical axis Conclusion In this paper, Optimal Homotopy Asymptotic Method (OHAM), Homotopy Perturbation Method (HPM) and Homotopy Analysis Method (HAM) have been used for numerical solution of Riccati Equation. We have compared the results of each of them with the exact solution of the given problem. Conclusions reveal that the method OHAM is more effective and the results obtained by this method are in good agreement with the exact solution. 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