SOLUCION DE SISTEMA DE ECUACIONES NO LINEALES POR EL METODO DE NEWTON

X=Xi+1=Xi - f dg – g df dy dy

df dg – df dg dx dy dy dx

Y=Yi+1=Yi - g df – f dg dx dx

df dg – df dg dx dy dy dx

v.vXY

Xi=1.5Yi=3.5

{2}{3}

f (X*Y) = X + XY – 10 = 0 2

2g (X*Y) = Y + 3XY – 57 = 0

f = (1.5) + (1.5)(3.5) – 10 = -2.5

g = (3.5) + 3 (1.5)(3.5) – 57= 1.6252

2

dfdx = 2X + Y = 6.5

dx

dydf = X = 1.5

dg= 3Y = 3.675

2

dgdy

=1 + 6XY = 32.5

1.5 – (-2.5)(32.5) – (1.625)(1.5)

(6.5)(32.5) – (1.5)(36.75)

Xi = 2.0360

3.5 – (1.625)(6.5) – (2.5)(36.75)

(6.5)(32.5) – (1.5)(36.75)

Yi = 2.8439

dfdx = 2X + Y = 6.9159

dx

dydf = X = 2.0360

dg= 3Y = 24.2633

2

dgdy

=1 + 6XY = 35.74

Xi = 2.0360Yi = 2.8439

f = (2.0360) + (2.0360)(2.8439) – 10= -0.06452

g = (2.8939) + 3(2.0360)(2.8439) – 57= -4.75602

X= 2.0360 – (-0.0645)(35.7410)-(-4.7560)(2.0360)

(6.9159)(35.74108)-(2.0360)(24.2633)

Xi=1.9986

Y= 2.8439 – (4.7560)(6.9159) – (-0.0645)(24.2633)

(6.9159)(35.74108)-(2.0360)(24.2633)

Yi=3.00229

Xi=1.9986

Yi=3.00229

dfdx = 2X + Y = 6.9984

dx

dydf = X = 1.9986

dg= 3Y = 27.0396

2

dgdy

=1 + 6XY = 37.0011

X= 1.9986 –(-0.0077990)(37.0011)-(0.0435)(1.9986)

f = (1.9986) + (1.9986)(3.00229) -10= - 0.00779902

g = ( 3 . 0 0 2 2 9 ) + 3 ( 1 . 9 9 8 6 ) ( 3 . 0 0 2 2 9 ) – 5 7 = 0 . 0 4 3 52

(6.9994)(37.0011) – (1.9986)(27.0396)

Xi=2.004

-0.3755 204.9441Xi=

Y=3.00229 – (0.0435)(6.9994) – (-0.0077990)(27.0396)

(6.9994)(37.0011) – (1.9986)(27.0396)

Yi=2.9994