Metodo heun

Ejemplo método de HEUN
Aplicar el método de HEUN al problema Y’= y-x2+1 Usando pasos de .2 H=.2Condición inicial: Cuando x0=0 y0=0.5Desde x=0 Hasta x=1

Formulas:


º 1ra ITERACIÒN
Y0i+1=0.5+f(0.5-(0)2+1).2=o.8y1=0.5+.2[(1.5)+(1.76)/2]=0.826
x1=.2
F(x1,y1 0)=0.86-(.2)2+1=1.786

º2da ITERACIÒN
y20=y1+h(f[x1,y1])=0.826+.2(1.786)=1.1832
x2=.4
F(x2,y2 0)=1.1832-(.4)2+1=2.0232
Y2=0.826+.2[(1.786)+(2.0232)/2]=1.20692
F(x2,y20)=1.20692-(.4)2+1=2.04692

º3ra ITERACIÒN
y30=y2+h(f[x2,y2])= 1.20692+.2(2.04692)=1.616304
x3=.6
F(x3,y30)=1.616304-(.6)2+1=2.256304
Y3=1.20692+.2[(2.256304)+(2.04692)/2]=1.6372424
F(x3,y3 0)=1.6372424-(.6)2+1=2.2772424

º4ta ITERACIÒNy40=y3+h(f[x3,y3])= 1.6372424+.2(2.2772424)=2.09269088
x4=.8
Y4=1.6372424+.2[(2.2772424)+(2.4529088)/2]=2.110237328F(x4,y40)= 2.110237328 -(.8)2+1=2.470237328

º5ta ITERACIÒN
y50=y4+h(f[x4,y4])=2.110237328+.2(2.470237328)=2.604284794
x5=1
Y5=2.110237328+.2[(2.470237328)+( 2.604284794)/2]=2.61768954
F(x5,y50)= 2.61768954-(1)2+1=2.61768954