Modified Euler's Method :

The Euler forward scheme may be very easy to implement but it can't give accurate solutions.    A  very small step size is required for any meaningful result.  In this scheme, since, the starting point of each sub-interval is used to find the slope of the solution curve,  the solution would be correct only if the function is linear. So an improvement over this is to take the arithmetic average of the slopes at x_i  and x_i+1(that is, at the end points of each sub-interval). The scheme so obtained is called modified Euler's method. It works first by approximating a value to y_i+1 and then improving it by making use of average slope.
 

yi+1	= yi+ h/2 (y'i + y'i+1)
	= yi + h/2(f(xi, yi) + f(xi+1, yi+1))

 If Euler's method is used to find the first approximation of y_i+1 then 

y_i+1 = y_i + 0.5h(f_i  + f(x_i+1, y_i + hf_i))

Truncation error:

y_i+1 = y_i + h y'_i  + h²y_i'' /2  + h³y_i''' /3! + h⁴y_i^iv /4! + . . .
f_i+1 = y'_i+1  =  y'_i + h y''_i  + h²y_i'''' /2  + h³y_i^iv /3! + h⁴y_i^v /4! + . . .

By substituting these expansions in the Modified Euler formula gives

y_i + h y'_i  + h²y_i'' /2  + h³y_i''' /3! + h⁴y_i^iv /4! + . . . = y_i+ h/2 (y'_i + y'_i + h y''_i  + 
                                                                                    h²y_i'''' /2  + h³y_i^iv /3! + h⁴y_i^v /4! + . . . )

So the truncation error is: - h³y_i''' /12 - h⁴y_i^iv /24 + . . .  that is, Modified Euler's method is of order two. 
 

Worked out problems

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 length 0.2.	Solution
Example 2	Find  y  in  [0,3] by solving the initial value problem y' = (x - y)/2,  y(0) = 1.  Compare solutions for h = 1/2, 1/4 and 1/8.	Solution
Example 3	Find y(0.1) for y' = x - y2,  y(0) = 1 correct upto four decimal places.	Solution
Example 4	Find y at x = 1.1 and 1.2 by solving y' = x2 + y2 ,  y(1) = 2.3	Solution

 
 

Problems to workout