Formula

6. Modified Euler method `y_(m+1)=y_m+hf(x_m+1/2 h,y_m + 1/2 hf(x_m,y_m))`

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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 Modified 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)`

Modified Euler method
`y_(m+1)=y_m+hf(x_m+1/2 h,y_m + 1/2 hf(x_m,y_m))`

`f(x_0,y_0)=f(0,1)=-0.5`

`x_0+1/2 h=0+0.1/2 =0.05`

`y_0 + 1/2 hf(x_0,y_0)=1+0.1/2 * -0.5=0.975`

`f(x_0+1/2 h, y_0 + 1/2 hf(x_0,y_0)=f(0.05,0.975)=-0.4625`

`y_1=y_0+hf(x_0+1/2 h, y_0 + 1/2 hf(x_0,y_0))=1+0.1*-0.4625=0.9538`

`:.y(0.1)=0.9538`

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Again taking `(x_1,y_1)` in place of `(x_0,y_0)` and repeat the process

`f(x_1,y_1)=f(0.1,0.9538)=-0.4269`

`x_1+1/2 h=0.1+0.1/2 =0.15`

`y_1 + 1/2 hf(x_1,y_1)=0.9538+0.1/2 * -0.4269=0.9324`

`f(x_1+1/2 h, y_1 + 1/2 hf(x_1,y_1)=f(0.15,0.9324)=-0.3912`

`y_2=y_1+hf(x_1+1/2 h, y_1 + 1/2 hf(x_1,y_1))=0.9538+0.1*-0.3912=0.9146`

`:.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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