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y 1= 2 .16667t 1= t 0+ h = 0 + 1 = 1Iteración2: y 2= y 1 +6h (k1 + 2k 2+2k 3+ k 4)k 1= f(t 1,y 1)= f(1; 2.16667 ) = 5.33333k 2= f(t 1+h/2,y 1+hk 1/2)= f(1.5; 4.83333)=16.k 3= f(t 1+h/2,y 1+h k 2/2)= f(1.5,10.1667)=32.k 4= f(t 1+h,y 1+ hk 3) =f (2; 34.1667) = 138.667y 2=2.16667+ (5.33333+ 216. + 2 32. + 138.667)= 42.1667t 2= t 1+ h = 1+ 1 = 2Por lo tanto, y(2) y(t 2)= 42.1667Finalmente, y(2) 42.16679Comparación las aproximaciones de y’ = 2 t y + t, con y( 0 )=0.5, en el intervalo[0,2], obtenidas con el método de Runge Kutta de orden 4 (clásico) paradiferentes tamaños de pasot yaprox yexacta0 0.5000 0.50001.0000 2.1667 2.21832.0000 42.1667 54.09820t yaprox yexacta0 0.5000 0.50000.2500 0.5645 0.56450.5000 0.7840 0.78400.7500 1.2550 1.25511.0000 2.2179 2.21831.2500 4.2685 4.27071.5000 8.9760 8.98771.7500 20.8212 20.88092.0000 53.7907 54.0982t yaprox yexacta0 0.5000 0.50000.5000 0.7839 0.78401.0000 2.2131 2.21831.5000 8.8724 8.98772.0000 51.5849 54.0982t yaprox yexacta0 0.5000 0.50000.1000 0.5101 0.51010.2000 0.5408 0.54080.3000 0.5942 0.59420.4000 0.6735 0.67350.5000 0.7840 0.78400.6000 0.9333 0.9333… ….. ….1.7000 17.4918 17.49331.8000 25.0307 25.03371.9000 36.4601 36.46612.0000 54.0863 54.0982105