Example 1:
Find   y(1.0)   accurate  upto  four decimal places using Modified Euler's method by solving the IVP  y' = -2xy²,  y(0) = 1 with step lengh 0.2.

Solution:        f(x, y) = -2xy²
y' = -2*x*y*y,  y[0.0] = 1.0 with h = 0.2

Given
y[0.0] = 1.0

Euler Solution:      y(1) = y(0) + h*(-2*x*y*y)(1)
y[0.20] = 1.0
Modified Euler iterations:y(1) = y(0) + .5*h*((-2*x*y*y)(0) + (-2*x*y*y)(1)
y[0.20] = 1.0   y[0.20] = 0.9599999988079071   y[0.20] = 0.9631359989929199   
y[0.20] = 0.9628947607919341   y[0.20] = 0.9629133460803093 

Euler Solution:      y(2) = y(1) + h*(-2*x*y*y)(2)
y[0.40] = 0.8887359638083165
Modified Euler iterations:y(2) = y(1) + .5*h*((-2*x*y*y)(1) + (-2*x*y*y)(2)
y[0.40] = 0.8887359638083165   y[0.40] = 0.8626358081578545   
y[0.40] = 0.8662926943348495   y[0.40] = 0.8657868947404332   
y[0.40] = 0.865856981554814 

Euler Solution:      y(3) = y(2) + h*(-2*x*y*y)(3)
y[0.60] = 0.7458966289094106
Modified Euler iterations:y(3) = y(2) + .5*h*((-2*x*y*y)(2) + (-2*x*y*y)(3)
y[0.60] = 0.7458966289094106   y[0.60] = 0.7391085349039348   
y[0.60] = 0.7403181774980547   y[0.60] = 0.7401034281837107   
y[0.60] = 0.7401415785278189 

Euler Solution:      y(4) = y(3) + h*(-2*x*y*y)(4)
y[0.80] = 0.6086629119889084
Modified Euler iterations:y(4) = y(3) + .5*h*((-2*x*y*y)(3) + (-2*x*y*y)(4)
y[0.80] = 0.6086629119889084   y[0.80] = 0.6151235687114084   
y[0.80] = 0.6138585343771569   y[0.80] = 0.6141072871136244   
y[0.80] = 0.6140584135348263 

Euler Solution:      y(5) = y(4) + h*(-2*x*y*y)(5)
y[1.00] = 0.49340256427369866
Modified Euler iterations:y(5) = y(4) + .5*h*((-2*x*y*y)(4) + (-2*x*y*y)(5)
y[1.00] = 0.49340256427369866   y[1.00] = 0.5050460713552334   
y[1.00] = 0.5027209825340415   y[1.00] = 0.5031896121302805   
y[1.00] = 0.5030953322323046   y[1.00] = 0.503114306721248