Euler

Post on 08-Jul-2016

220 views 0 download

description

Ejercicios metodo de Euler

Transcript of Euler

y '=x+ y , y (0 )=2, n=10, h=0.1

y1= y0+hF (x0 , y0 )=2+0.1 (0+2 )=2.2

y2= y1+hF (x1 , y1 )=2 .2+0.1 (0.1+2 .2 )=2.43

y3= y2+hF (x2 , y2 )=2 .43+0.1 (0 .2+2.43 )=2.693

y4= y3+hF (x3 , y3 )=2 .693+0.1 (0 .3+2.693 )=2.9923

y5= y 4+hF (x4 , y4 )=2 .9923+0.1 (0.4+2.9923 )=3.33153

y6= y5+hF (x5 , y5 )=3.33153+0.1 (0.5+3.33153 )=3.714683

y7= y6+hF (x6 , y6 )=3.714683+0.1 (0 .6+3.714683 )=4.1461513

y8= y7+hF (x7 , y7 )=4.1461513+0.1 (0 .7+4.1461513 )=4.63076643

y9= y8+hF (x8 , y8 )=4.63076643+0.1 (0 .8+4.63076643 )=¿5.17384307

y10= y9+hF (x9 , y9 )=5.17384307+0.1 (0 .9+5.17384307 )=¿5.78122738

n 0 1 2 3 4 5 6 7 8 9 10xn 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1yn 2 2.2 2.43 2.69

32.9923 3.33153 3.714683 4.1461513 4.63076643 5.17384307 5.781227377

y '=3 x−2 y , y (0 )=3, n=10, h=0.05

y1= y0+hF (x0 , y0 )=3+0.05 (3(0)+2(3))=¿2.7

y2= y1+hF (x1 , y1 )=2.7+0.05 (3 (0.05 )−2(2.7))=2.4375

y3= y2+hF (x2 , y2 )=2.4375+0.05 (3 (0.10 )−2(2.4375))=2.20875

y4= y3+hF (x3 , y3 )=2.20875+0.05 (3 (0.15 )−2(2.20875))=2.010375

y5= y 4+hF (x4 , y4 )=2.010375+0.05 (3 (0 .20 )−2(2.010375))=1.8393375

y6= y5+hF (x5 , y5 )=1.8393375+0.05 (3 (0.25 )−2(1.8393375))=1.69290375

y7= y6+hF (x6 , y6 )=1.69290375+0.05 (3 (0.30 )−2(1.69290375))=1.56861338

y8= y7+hF (x7 , y7 )=1.56861338+0.05 (3 (0.35 )−2(1.56861338))=1.46425204

y9= y8+hF (x8 , y8 )=1.46425204+0.05 (3 (0.40 )−2(1.46425204))=1.37782684

y10= y9+hF (x9, y9 )=1.37782684+0.05 (3(0.45)+2(1.37782684 ))=1.30754416

n 0 1 2 3 4 5 6 7 8 9 10

xn 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

yn 3 2.7 2.4375 2.20875 2.010375

1.8393375

1.69290375

1.56861338

1.46425204

1.37782684

1.30754416

y '=0.5 x (3− y) , y (0 )=1, n=5, h=0.4

y1= y0+hF (x0 , y0 )=1+0.4¿1

y2= y1+hF (x1 , y1 )=1+0.4 ¿

y3= y2+hF (x2 , y2 )=1 .16+0.4 ¿

y4= y3+hF (x3 , y3 )=1.4544+0.4¿

y5= y 4+hF (x4 , y4 )=1.825344+0.4¿

n 0 1 2 3 4 5

xn 0 0.4 0.8 1.2 1.6 2

yn 1 1 1.16 1.4544 1.825344 2.20123392

y '=e xy , y (0 )=1, n=10, h=0.1

y1= y0+hF (x0 , y0 )=1+0.1 (e(0)(1 ))=1.1

y2= y1+hF (x1 , y1 )=1 .1+0.1 (e(0.1)(1 .1))=1.21162781

y3= y2+hF (x2 , y2 )=1.211627807+0.1 (e(0 .2)(1.211627807))=¿1.3390487

y4= y3+hF (x3 , y3 )=1.3390487+0.1 (e(0 .3)(1.3390487))=1.48848718

y5= y 4+hF (x4 , y4 )=1.48848718+0.1 (e(0 .4)(1.48848718))=1.66986188

y6= y5+hF (x5 , y5 )=1.66986188+0.1(e(0 .5)(1.66986188))=1.90032737

y7= y6+hF (x6 , y6 )=1.90032737+0.1 (e(0 .6)(1.90032737))=¿2.21306563

y8= y7+hF (x7 , y7 )=2.21306563+0.1 (e(0 .7)(2.21306563))=¿2.68381043

y9= y8+hF (x8 , y8 )=2.68381043+0.1 (e(0 .8)(2.68381043))=¿3.53976605

y10= y9+hF (x9, y9 )=3.53976605+0.1 (e(0 .9)(3.53976605))=¿5.95840352

n 0 1 2 3 4 5 6 7 8 9 10

xn 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

yn 1 1.1 1.21162781

1.3390487

1.48848718

1.66986188

1.90032737

2.21306563

2.68381043

3.539766051

5.95840352

y '=cosx+seny , y (0 )=5, n=10, h=0.1

y1= y0+hF (x0 , y0 )=5+0.1 (cos (0 )+sen (5))=5.108715574

y2= y1+hF (x1 , y1 )=5.108715574+0.1 (cos (0.1 )+sen (5.108715574))=5.217620003

y3= y2+hF (x2 , y2 )=5.217620003+0.1 (cos (0.2 )+sen (5.217620003))=¿5.326713277

y4= y3+hF (x3 , y3 )=5.326713277+0.1 (cos (0.3 )+sen (5.326713277))=5.435995381

y5= y 4+hF (x4 , y4 )=5.435995381+0.1 (cos (0.4 )+sen (5.435995381))=5.545466319

y6= y5+hF (x5 , y5 )=5.545466319+0.1 (cos (0.5 )+sen(5.545466319))=¿5.655126072

y7= y6+hF (x6 , y6 )=5.655126072+0.1 (cos (0.6 )+sen(5.655126072))=¿5.764974628

y8= y7+hF (x7 , y7 )=5.764974628+0.1 (cos (0.7 )+sen (5.764974628))=5.875011975

y9= y8+hF (x8 , y8 )=5.875011975+0.1 (cos (0.8 )+sen(5.875011975))=5.985238099

y10= y9+hF (x9 , y9 )=5.985238099+0.1 (cos (0.9 )+sen(5.985238099))=6.095652985

n 0 1 2 3 4 5 6 7 8 9 10

xn 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

yn 5 5.108715574

5.217620003

5.32671

3277

5.435995381

5.545466319

5.655126072

5.764974628

5.875011975

5.985238099

6.095652985