Formula

5. Improved Euler method `y_(m+1)=y_m+1/2 h[f(x_m,y_m) + f(x_m+h,y_m + 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 Improved 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)`

Improved Euler method
`y_(m+1)=y_m+1/2 h[f(x_m,y_m) + f(x_m+h,y_m + hf(x_m,y_m))]`

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

`f(x_0+h,y_0 + hf(x_0,y_0))=f(0.1,0.95)=-0.425`

`y_1=y_0+1/2 h[f(x_0,y_0) + f(x_0+h,y_0 + hf(x_0,y_0))]`

`y_1=1+0.1/2 * [-0.5-0.425]=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`

`f(x_1+h,y_1 + hf(x_1,y_1))=f(0.2,0.9111)=-0.3555`

`y_2=y_1+1/2 h[f(x_1,y_1) + f(x_1+h,y_1 + hf(x_1,y_1))]`

`y_2=0.9538+0.1/2 * [-0.4269-0.3555]=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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