Numerical Methods for Ordinary Dierential Equations Episode 1 docx

... examples 10 3 16 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS λ y 1 y 2 y 3 y λ x 0 0.2 0.5 1 2 5 10 20 50 10 2 10 3 10 4 0 10 −5 10 −4 10 −3 10 −2 10 1 0.5 1. 0 −0.0 10 4 −0.2 10 4 −0.4 10 4 −0.6 10 4 −0.8 10 4 1. 0 10 4 Figure ... equations dy 1 dx = y 3 , (10 1a) dy 2 dx = y 4 , (10 1b) dy 3 dx = − y 1 (y 2 1 + y 2 2 ) 3/2 , (10 1...

Ngày tải lên: 13/08/2014, 05:21

... θ =1, A2 n + B +2Cn2 n ,θ=2, A2 n + B + Cθ 2 (θ 1) (θ−2) θ n ,θ =1, θ=2. 32 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS 0 1 2 3 4 √ 2 1+ √ 2 2 √ 2 2+ √ 2 1+ 2 √ 2 10 10 2 1 Figure 12 1(ii) ... polynomial equations. Consider the difference equations y n = y n 1 − 1 2 y 2 n 1 +1, y 0 =0, (13 4a) y n = y n 1 − 1 2 y 2 n 1 +1, y 0 =4, (13 4b) y n = y n...

Ngày tải lên: 13/08/2014, 05:21

... is              000 − 529 3375 3904 3375 4232 3375 14 72 3375 18 9 92 00 15 2 25 − 12 7 25 − 419 10 0 − 11 18 575 3375 515 2 25 16 8 0 10 19 96 − 1 552 3375 515 2 25 16 8 0 10 19 96 − 1 552 000 10 00 0 01 0000 000 0 010              . For ... 10 −5 4.58854 82 8 −9 1. 02365 × 10 −5 4.78350 13 7 8 10 2 .15 123 × 10 −6 4.75845 228 8...

Ngày tải lên: 13/08/2014, 05:21

... 11 30400.0252 10 10 4. 412 5 40 25 617 8.9889 10 10 4 .18 93 80 611 50.2626 10 10 4.0904 16 0 14 949. 617 6 10 10 4.0442 320 3696.5967 10 10 4.0 218 640 919 .13 62 10 10 4. 010 8 12 80 229 .16 29 10 10 4.0054 2560 ... 0.0000 0. 01 1. 01 0. 010 2 0. 010 2 1. 02 0.0002 0.02 1. 02 0. 010 4 0. 010 6 1. 03 0.0006 0.03 1. 03 0. 010 6 0. 011 2 1. 04 0.0 012 0.04 1....

Ngày tải lên: 13/08/2014, 05:21

... − 1 2 2 3 2 − 1 2 5 12 3 23 12 − 4 3 5 12 − 3 8 4 55 24 − 59 24 37 24 − 3 8 2 51 720 5 19 01 720 − 13 87 360 10 9 30 − 637 360 2 51 720 − 95 288 6 4277 14 40 − 26 41 480 49 91 720 − 3649 720 959 480 − 95 288 19 087 60480 7 19 87 21 60480 − 18 637 2520 23 518 3 2 016 0 − 10 754 945 13 5 713 2 016 0 − 5603 2520 19 087 60480 − 5257 17 280 8 16 083 4480 − 11 5 216 9...

Ngày tải lên: 13/08/2014, 05:21

... c 3 −4 2 (1 c 3 )(6c 3 1 6c 2 3 ) c 3 (1 2c 3 ) 2 (1 c 3 )(6c 3 1 6c 2 3 ) 1 c 3 6c 3 1 6c 2 3 6c 3 1 6c 2 3 12 c 3 (1 c 3 ) 1 12c 3 (1 c 3 ) 1 12c 3 (1 c 3 ) 6c 3 1 6c 2 3 12 c 3 (1 c 3 ) ; (322e) for case II, 0 c 2 c 2 1 2 1 2 − 1 8c 2 1 8c 2 1 1 2c 2 − 1 − 1 2c 2 2 1 6 0 2 3 1 6 ; (322f) for case ... c 3 =0,b 3 =0. 18 0 NUMERICAL METHODS FOR...

Ngày tải lên: 13/08/2014, 05:21

... method 0 1 2 1 2 1 2 1 4 1 4 7+ √ 21 14 1 7 −7−3 √ 21 98 21+ 5 √ 21 49 7+ √ 21 14 11 + √ 21 84 0 18 +4 √ 21 63 21 √ 21 252 1 2 5+ √ 21 48 0 9+ √ 21 36 −2 31+ 14 √ 21 360 63−7 √ 21 80 7− √ 21 14 10 − √ 21 42 0 −432+92 √ 21 315 633 14 5 √ 21 90 −504 +11 5 √ 21 70 63 13 √ 21 35 7− √ 21 14 1 14 00 0 14 −3 √ 21 126 13 −3 √ 21 63 1...

Ngày tải lên: 13/08/2014, 05:21

... 3   l m 01 2 3 0 11 +z 1+ z+ 1 2 z 2 1+ z+ 1 2 z 2 + 1 6 z 3 1 1 1 z 1+ 1 2 z 1 1 2 z 1+ 2 3 z+ 1 6 z 2 1 1 3 z 1+ 3 4 z+ 1 4 z 2 + 1 24 z 3 1 1 4 z 2 1 1−z+ 1 2 z 2 1+ 1 3 z 1 2 3 z+ 1 6 z 2 1+ 1 2 z+ 1 12 z 2 1 1 2 z+ 1 12 z 2 1+ 3 5 z+ 3 20 z 2 + 1 60 z 3 1 2 5 z+ 1 20 z 2 3 1 1−z+ 1 2 z 2 − 1 6 z 3 1+ 1 4 z 1 3...

Ngày tải lên: 13/08/2014, 05:21

... conditions c 1 = α 1 + a 11 + a 12 , 1 2 c 2 1 = α 2 + a 11 c 1 + a 12 c 2 , c 2 = α 1 + a 21 + a 22 , 1 2 c 2 2 = α 2 + a 21 c 1 + a 22 c 2 , 1+ α 1 = α 1 + b 1 + b 2 , 1 2 + α 1 + α 2 = α 2 + b 1 c 1 + ... 2α 1 α 6 +2α 2 α 4 −α 2 1 α 3 −2α 2 1 α 4 −3α 1 α 2 2 +4α 3 1 α 2 −α 5 1 −α 13 14 5 3α 1 α 7 + α 1 α 5 − 3α 2 1 α 4 − 3α 2...

Ngày tải lên: 13/08/2014, 05:21

... β(t)(β 1 E)(t)(β 1 Eβ (r) )(t) 11 0 1 1 1 2 1 2 0 1 2 1 2 1 3 1 3 0 1 3 1 3 1 6 1 6 1 72 11 72 11 +r 3 72 1 4 1 4 1 108 13 54 26+r 4 10 8 1 8 5 36 1 216 13 10 8 26+3r 3 +r 4 216 1 12 1 9 − 1 216 19 216 19 +6r 3 −r 4 216 1 24 1 24 0 1 36 2+r 3 72 We ... β 14 β 14 +3β 7 +3β 4 + β 2 + 1 20 15 5 α 15 + β 2 α 4 ++β 6 α 1 +...

Ngày tải lên: 13/08/2014, 05:21