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12087524
MÉTODO DE RUNGE KUTTA yi+1=yi+∅(xi;yih)h
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Mind Map on MÉTODO DE RUNGE KUTTA yi+1=yi+∅(xi;yih)h, created by diego tuapante farez on 02/02/2018.
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diego tuapante farez
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Resource summary
MÉTODO DE RUNGE KUTTA yi+1=yi+∅(xi;yih)h
Método de Euler yi+1=yi+f(xi;yi)h
Error de redondeo
Error de truncamiento
Local
Propagado
Mejoras del Método de Euler
Método de Heun y´i+1=f(xi+1;yᵒi+1)
Predicador yi+1=yi+f(xi;yi)h
Corredor yi+1=(yi+f(xi;yi)+f(xi+1;y˚i+1)h)/2
Método punto medio o poligono mejorado yi+1/2=yi+f(xi;yi) h/2
Método de Runge Kutta yi+1=yi+∅(xi;yih)h
Método de Runge kutta segundo orden yi+1=yi+(a1k1+a2k2)h k1=f(xi;yi) k2=f(xi+p1h;yi+q11k1h) a1+a2=1 a2p1=1⁄2 a2q11=1⁄2
Método de Heun con un solo corrector a2=1⁄2 yi+1=yi+(1⁄2 k1+1⁄2 k2)h k1=f(xi;yi) k2=f(xi+h;yi+k1h)
Método del punto medio a2=1 yi+1=yi+k2h k1=f(xi;yi) k2=f(xi+1⁄2 h;yi+1⁄2 kh)
Método de Ralston a2=2⁄3 yi+1=(yi+1⁄3 k1+2⁄3 k2)h k1=f(xi;yi) k2=f(xi+3⁄4 h;yi+3⁄4 k1h)
Método de Runge kutta de tercer orden yi+1=yi+1⁄6 (k1+4k2+k3)h
k1=f(xi;yi) k2=f(xi+1⁄2 h;yi+1⁄2 k1h) k3=f(xi+h;yi-k1h+2k2h)
Método de runge kutta de cuarto orden yi+1=yi+1⁄6 (k1+2k2+2k3+k4)h
k1=f(xi;yi) k2=f(xi+1⁄2 h;yi+1⁄2 k1h) k3=f(xi+1⁄2 h;yi+1⁄2 k2h) k4=f(xi+h;yi+k3h)
Método de Runge kutta de orden superior yi+1=yi+1⁄90 (7k1+32k3+12k4+32k5+7k6)h
k1=f(xi;yi) k2=f(xi+1⁄4 h;yi+1⁄4 k1h) k3=f(xi+1⁄4 h;yi+1⁄8 k1h+1⁄8 k2h) k4=f(xi+1⁄2 h;yi-1⁄2 k2h+k3h) k5=f(xi+3⁄4 h;yi+3⁄16 k1h+9⁄16 k4h) k6=f(xi+h;yi-3⁄7 k1h+2⁄7 k2h+12⁄7 k3h-12⁄7 k4h+8⁄7 k5h)
Sistema de ecuaciones
dy1/dx=f1(x,y1,y2,…..,yn dy2/dx=f2(x,y1,y2,…..,yn ------------------------------------- -------------------------------------- ------------------------------------- dyn/dx=fn(x,y1,y2,…..,yn
Métodos adaptativos de Runge Kutta dy/dx=f(x)
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