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Example 4:
Find y at x = 1.1 and 1.2 by solving y' = x2 + y2 ,  y(1) = 2.3

Solution:
Given y' = f(x,y) = x2 + y2      Þ      y''   = 2x+2yy',               y''' = 2 + 2y'2 + 2yy''

At x = 1.0 and y = 2.3 with h = .1

y'   = (1.0)2 + (2.3)2   = 6.29
y''  = 2*(1.0)+2*(2.3)*(6.29) = 30.934
y''' = 2 + 2*(6.29)2 + 2*(6.29)*(30.934) = 223.4246

y(x0+h) = y(x0) + h y'(x0, y0) + h2  y''(x0, y0)/2! + h3 y'''(x0, y0)/3!  + . . .

           = 2.3 + (0.1) * 6.29 + (0.1)2  * 30.934/2! + (0.1)3 * 223.4246/3! +  o(h4)

           = 3.1209

 Þ   y1 = y(1.1) = 3.1209

Now at x = 1.1 and y = 3.1209 with h = 0.1

y'   = (1.1)2 + (3.1209)2   = 10.95
y''  = 2*(1.1)+2*(3.1209)*(10.95) = 70.5477
y''' = 2 + 2*(10.95)2 + 2*(10.95)*(70.5477) = 682.1496

y(x1+h) = y(x1) + h y'(x1, y1) + h2  y''(x1, y1)/2! + h3 y'''(x1, y1)/3!  + . . .

           = 3.1209 + (0.1) * 10.95 + (0.1)2  * 70.5477/2! + (0.1)3 * 682.1496/3! +  o(h4)

           = 4.6823

 Þ   y2 = y(1.2) = 4.6823