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January 30, 2017 | Author: GopejuGomezPerez | Category: N/A
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Es probable que el procedimiento más difundido y a la vez el más exacto para obtener soluciones a al problema de valor inicial y’ = f(x, y), si se conoce que y(x0) = y0 , sea el Método de Runge-Kutt orden. Para obtener un nuevo valor de y se usa la siguiente: Fórmula del Método de Runge-Kutta de cuarto orden.

y n1 = y n  Donde:

1  k 1 2k 2 2k 3  k 4  6

k 1 =hf  x n , y n 

1 1 h , y n k 1  2 2 1 1 k 3 =hf  x n  h , y n  k 2  2 2 k 4 =hf  x n h , y n k 3  k 2 =hf  x n 

y a la vez el más exacto para obtener soluciones aproximadas noce que y(x0) = y0 , sea el Método de Runge-Kutta de cuarto a siguiente: de Runge-Kutta de cuarto orden.

Método de Runge Kutta Ecuación diferencial

Solución particular: y(1) = 1

y' = 2xy a)

Solución x0 = y0 = h=

n 0 1 2 3 4 5

X 1.000000 1.100000 1.200000 1.300000 1.400000

1.000000 1.000000 0.100000 Y 1.000000 1.233674 1.552695 1.993687 2.611633

k1 0.200000 0.271408 0.372647 0.518359 0.731257

k2 0.231000 0.314957 0.434755 0.608274 0.863406

k3 0.234255 0.319965 0.442518 0.620412 0.882567

1.500000 3.49021064 y(1.5) = 3.49034296

a)

Solución x0 = y0 = h=

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

X 1.000000 1.050000 1.100000 1.150000 1.200000 1.250000 1.300000 1.350000 1.400000 1.450000

1.000000 1.000000 0.050000 Y 1.000000 1.107937 1.233678 1.380574 1.552706 1.755053 1.993714 2.276180 2.611692 3.011680

k1 0.100000 0.116333 0.135705 0.158766 0.186325 0.219382 0.259183 0.307284 0.365637 0.436694

k2 0.107625 0.125356 0.146422 0.171545 0.201619 0.237755 0.281338 0.334101 0.398218 0.476429

k3 0.108016 0.125841 0.147025 0.172296 0.202556 0.238926 0.282806 0.335944 0.400539 0.479359

1.500000 3.49033382 y(1.5) = 3.49034296

k4 0.271536 0.372873 0.518756 0.731948 1.048260 REAL

k4 0.116342 0.135716 0.158781 0.186344 0.219408 0.259217 0.307330 0.365697 0.436774 0.523656 REAL

Método de Runge Kutta Ecuación diferencial

x 2 y '  xy=1 a)

Solución x0 = 1.000000 y0 = 1.000000

Solución general

y=

ln x c  x x

y'=

Solución particular:

y=

1ln x x

1− x y 2 x

h = 0.100000 n 0 1 2 3 4 5

X 1.000000 1.100000 1.200000 1.300000 1.400000

Y 1.000000 0.995737 0.985268 0.971050 0.954623

k1 0.000000 -0.007877 -0.012661 -0.015525 -0.017167

k2 -0.004535 -0.010629 -0.014315 -0.016485 -0.017682

1.500000 0.93697685 y(1.5) =

b)

Solución x0 = 1.000000 y0 = 1.000000

y'=

1− x y x2

h = 0.050000 n 0 1 2 3 4 5 6 7 8 9 10

X 1.000000 1.050000 1.100000 1.150000 1.200000 1.250000 1.300000 1.350000 1.400000 1.450000 1.500000

Y 1.000000 0.998848 0.995737 0.991097 0.985268 0.978515 0.971049 0.963040 0.954623 0.945906 0.93697675

k1 0.000000 -0.002213 -0.003938 -0.005284 -0.006331 -0.007141 -0.007762 -0.008233 -0.008583 -0.008836

k2 -0.001190 -0.003140 -0.004661 -0.005846 -0.006766 -0.007476 -0.008017 -0.008424 -0.008722 -0.008933 y(1.5) =

ción particular: y(1) = 1

y=

1ln x x

k3 -0.004319 -0.010509 -0.014249 -0.016449 -0.017664

k4 -0.007872 -0.012658 -0.015522 -0.017165 -0.018020

0.93697674 REAL

k3 -0.001161 -0.003118 -0.004645 -0.005834 -0.006757 -0.007469 -0.008012 -0.008420 -0.008720 -0.008931

k4 -0.002212 -0.003938 -0.005284 -0.006330 -0.007140 -0.007762 -0.008233 -0.008583 -0.008836 -0.009010

0.93697674 REAL

Método de Runge Kutta Ecuación diferencial

y '  y=e 3x a)

Solución x0 = 0.000000 y0 = 1.000000

Solución general

y=

3 x

e 4



c ex

Solución particular: 1 y = [ e 3x 3 e− x 4

y ' =e 3x − y

h = 0.100000 n 0 1 2 3 4 5

X 0.000000 0.100000 0.200000 0.300000 0.400000

Y 1.000000 1.016094 1.069580 1.170519 1.332776

0.500000

1.575330

k1 0.000000 0.033376 0.075254 0.128908 0.198734

k2 0.016183 0.053553 0.100979 0.162268 0.242528 y(1.5) =

b)

Solución x0 = 0.000000 y0 = 1.000000

y ' =e 3x − y

h = 0.050000 n 0 1 2 3 4 5 6 7 8 9 10

X 0.000000 0.050000 0.100000 0.150000 0.200000 0.250000 0.300000 0.350000 0.400000 0.450000

Y 1.000000 1.003881 1.016093 1.037609 1.069578 1.113351 1.170515 1.242929 1.332770 1.442578

k1 0.000000 0.007898 0.016688 0.026535 0.037627 0.050182 0.064454 0.080736 0.099367 0.120742

k2 0.003894 0.012225 0.021528 0.031979 0.043782 0.057172 0.072421 0.089846 0.109812 0.132745

0.500000 1.57532090 y(1.5) =

ción particular: y(0) = 1 1 y = [ e 3x 3 e− x ] 4

k3 0.015374 0.052544 0.099693 0.160600 0.240339

k4 0.033448 0.075348 0.129033 0.198900 0.290857

1.57532026 REAL

k3 0.003797 0.012116 0.021407 0.031843 0.043628 0.056997 0.072222 0.089618 0.109551 0.132445

k4 0.007902 0.016693 0.026541 0.037633 0.050190 0.064463 0.080746 0.099378 0.120755 0.145333

1.57532026 REAL

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