Example 5:
Find y(0.8) with h = 0.1 from y' = y - 2x/y, y(0)
= 1 using Euler's method
Solution:
f(x, y) = y - 2x/y
y1 = y0 + h f(x0, y0) =
1.0 + 0.1* ( 1.0- 2*0.0/1.0 ) = 1.1000
y2 = y1 + h f(x1, y1) =
1.1 + 0.1* ( 1.1- 2*0.1/1.1 ) = 1.1918
y3 = y2 + h f(x2, y2) =
1.1918 + 0.1* ( 1.1918- 2*0.2/1.1918 ) = 1.2774
y4 = y3 + h f(x3, y3) =
1.2774 + 0.1* ( 1.2774- 2*0.3/1.2774 ) = 1.3582
y5 = y4 + h f(x4, y4) =
1.3582 + 0.1* ( 1.3582- 2*0.4/1.3582 ) = 1.4351
y6 = y5 + h f(x5, y5) =
1.4351 + 0.1* ( 1.4351- 2*0.5/1.4351 ) = 1.5089
y7 = y6 + h f(x6, y6) =
1.5089 + 0.1* ( 1.5089- 2*0.6/1.5089 ) = 1.5803
y8 = y7 + h f(x7, y7) =
1.5803 + 0.1* ( 1.5803- 2*0.7/1.5803 ) = 1.6497
