Formula

Midpoint Euler method `y_(n+1)=y_n+hf(x_n+h/2 ,y_n + h/2 f(x_n,y_n))` Or `y_(n+1)=y_n+h k_(2y)` `k_(2y)=f(x_n+h/2,y_n+h/2 k_(1y))` `k_(1y)=f(x_n,y_n)`

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Examples
1. Find y(0.2) for `y'=(x-y)/2`, `x_0=0, y_0=1`, with step length 0.1 using Midpoint Euler method (1st order derivative)

Solution:
Given `y'=(x-y)/(2), y(0)=1, h=0.1, y(0.2)=?`

Here, `x_0=0,y_0=1,h=0.1,x_n=0.2`

`y'=(x-y)/(2)`

`:. f(x,y)=(x-y)/(2)`

Midpoint Euler method
`y_(n+1)=y_n+hf(x_n+h/2 ,y_n + h/2 f(x_n,y_n))`

Or
`y_(n+1)=y_n+h k_(2y)`

`k_(2y)=f(x_n+h/2,y_n+h/2 k_(1y))`

`k_(1y)=f(x_n,y_n)`

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for `n=0,x_0=0,y_0=1`

`k_(1y)=f(x_0,y_0)`

`=f(0,1)`

`=-0.5`

`k_(2y)=f(x_0+h/2,y_0+h/2 k_(1y))`

`=f(0+0.1/2,1+0.1/2 *-0.5)`

`=f(0.05,0.975)`

`=-0.4625`

`y_(1)=y_0+h k_(2y)`

`=1+0.1*-0.4625`

`=0.9538`

`x_1=x_0+h=0+0.1=0.1`

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for `n=1,x_1=0.1,y_1=0.9538`

`k_(1y)=f(x_1,y_1)`

`=f(0.1,0.9538)`

`=-0.4269`

`k_(2y)=f(x_1+h/2,y_1+h/2 k_(1y))`

`=f(0.1+0.1/2,0.9538+0.1/2 *-0.4269)`

`=f(0.15,0.9324)`

`=-0.3912`

`y_(2)=y_1+h k_(2y)`

`=0.9538+0.1*-0.3912`

`=0.9146`

`x_2=x_1+h=0.1+0.1=0.2`

`:.y(0.2)=0.9146`

`n`	`x_n`	`y_n`	`x_(n+1)`	`y_(n+1)`
0	0	1	0.1	0.9538
1	0.1	0.9538	0.2	0.9146

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