1. Formula & Example-1
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`, y(0) = 1, with step length 0.1 using Modified 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`
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=0.95375`
`:.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`
`x_1+1/2 h=0.1+0.1/2 =0.15`
`y_1 + 1/2 hf(x_1,y_1)=0.95375+0.1/2 * -0.42688=0.93241`
`f(x_1+1/2 h, y_1 + 1/2 hf(x_1,y_1)=f(0.15,0.93241)=-0.3912`
`y_2=y_1+hf(x_1+1/2 h, y_1 + 1/2 hf(x_1,y_1))=0.95375+0.1*-0.3912=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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