1. Dada la ecuación diferencial: y'=x2+y2 y(2)=0.5
Usa el método Euler mejorado para aproximar y(2.3) tomando h=0.1 en cada paso del proceso iterativo. 1ra iteración: x1=x0+h=2+0.1=2.1 y1*=y0+(h(f(x0,y0))=0.5+(0.1(2.06155))=0.70616 y1=y0+h(fx0,yo+fx1,y1*2)=0.5+0.1(22+0.52+2.12+0.7061622) =0.5+0.1(2.06155+2.215552)=0.71386 2da iteración: x2=x1+h=2.1+0.1=2.2 y2*=y1+(h(f(x0,y0))=0.71386+(0.1(2.06155))=0.92002 y2=y1+h(fx1,y1+fx2,y2*2)
=0.71386+0.1(2.12+0.713862+2.22+0.9200222)
=0.71386+0.1(2.21802+2.384632)=0.94399
3ra iteración: x3=x2+h=2.2+0.1=2.3 y3*=y2+(h(f(x0,y0))=0.94399+(0.1(2.06155))=1.15015 y3=y2+h(fx2,y2+fx3,y3*2)
=0.94399+0.1(2.22+0.943992+2.32+1.1501522) …ver más…

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=10.49469+0.1(3.1+10.49469+(3.2+10.98412)2)
=10.49469+0.1(5.00023+5.103082)=10.99986
3ra iteración: x3=x2+h=3.2+0.1=3.3 y3*=y2+(h(f(x0,y0))=10.99986+(0.1(4.89433))=11.48929 y3=y2+h(fx2,y2+fx3,y3*2)
=10.99986+0.1(3.2+10.99986+(3.3+11.48929)2)
=10.99986+0.1(5.10546+5.206182)=11.51544 n | Xn | Yn* | Yn | 0 | 3 | -------- | 10 | 1 | 3.1 | 10.48943 | 10.49469 | 2 | 3.2 | 10.98412 | 10.99986 | 3 | 3.3 | 11.48929 | 11.51544