Example 2:
Use Eulers method to solve for y[0.1] from y' = x + y + xy,  y(0) = 1  with  h = 0.01 also estimate how small  h would need to obtain four-decimal accuracy.

Solution :
f(x, y) = x + y + xy,

y₁ = y₀ + h f(x₀, y₀) = 1.0 + .01*(0 + 1 + 0*1) = 1.01
y₂ = y₁ + h f(x₁, y₁) = 1.01 + .01*(0.01 + 1.01 + 0.01*1.01) =1.02
y₃ = y₂ + h f(x₂, y₂) = 1.02 + .01*(0.02 + 1.02 + 0.02*1.02) =1.031
y₄ = y₃ + h f(x₃, y₃) = 1.031 + .01*(0.03 + 1.031 + 0.03*1.031) =1.042
y₅ = y₄ + h f(x₄, y₄) = 1.042 + .01*(0.04 + 1.042 + 0.04*1.042) = 1.053
y₆ = y₅ + h f(x₅, y₅) = 1.053 + .01*(0.05 + 1.053 + 0.05*1.053) = 1.065
y₇ = y₆ + h f(x₆, y₆) = 1.065 + .01*(0.06 + 1.065 + 0.06*1.065) = 1.076
y₈ = y₈ + h f(x₇, y₇) = 1.076 + .01*(0.07 + 1.076 + 0.07*1.076) = 1.089
y₉ = y₉ + h f(x₈, y₈) = 1.089 + .01*(0.08 + 1.089 + 0.08*1.089) = 1.101
y₁₀ = y₁₀ + h f(x₉, y₉) = 1.101 + .01*(0.09 + 1.101 + 0.09*1.101) = 1.114

error in the approximation = ( h²/2 )  f''(x),    where 0 < x <0.1
                                        = 0.00005 f''(x)