1. Formula & Example-1
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`, y(0) = 1, with step length 0.1 using Improved Euler method
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`
`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]=0.95375`
`:.y(0.1)=0.95375`
Again taking `(x_1,y_1)` in place of `(x_0,y_0)` repeat the process
`f(x_1,y_1)=f(0.1,0.95375)=-0.42688`
`f(x_1+h,y_1 + hf(x_1,y_1))=f(0.2,0.91106)=-0.35553`
`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=0.95375+0.1/2 * [-0.42688-0.35553]=0.91463`
`:.y(0.2)=0.91463`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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