1. Formula & Example-1 `y'=(x+y)/2` (table data)
Formula
Milne's simpson predictor formula is
`y_(n+1,p) = y_(n-3) + (4h)/3 (2y'_(n-2) - y'_(n-1)+ 2y'_(n))`
putting `n=3`, we get
`y_(4,p)=y_0 + (4h)/3 (2y'_1 - y'_2 + 2y'_3)`
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Milne's simpson corrector formula is
`y_(n+1,c) = y_(n-1) + h/3 (y'_(n-1) + 4y'_(n) + y'_(n+1))`
putting `n=3`, we get
`y_(4,c) = y_2 + h/3 (y'_2 + 4y'_3 + y'_4)`
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Examples
1. `y'=(x+y)/2`,
| `x_i` | 0 | 0.5 | 1 | 1.5 | | `y_i` | 2 | 2.636 | 3.595 | 4.968 |
Find y(2) by Milne's simpson predictor corrector method
Solution:
`y'=(x+y)/2`
Milne's simpson predictor formula is
`y_(n+1,p) = y_(n-3) + (4h)/3 (2y'_(n-2) - y'_(n-1)+ 2y'_(n))`
putting `n=3`, we get
`y_(4,p)=y_0 + (4h)/3 (2y'_1 - y'_2 + 2y'_3) ->(2)`
We have given that
`x_0=0,x_1=0.5,x_2=1,x_3=1.5`
and using runge kutta 4 method, we get
`y_0=2,y_1=2.636,y_2=3.595,y_3=4.968`
`y'=(x+y)/2`
`y'_1=(x+y)/2=1.568` (where `x=0.5,y=2.636`)
`y'_2=(x+y)/2=2.2975` (where `x=1,y=3.595`)
`y'_3=(x+y)/2=3.234` (where `x=1.5,y=4.968`)
putting the values in (2), we get
`y_(4,p)=y_0 + (4h)/3 (2y'_1 - y'_2 + 2y'_3) ->(2)`
`y_(4,p)=2 + (4*0.5)/3 * (2 * 1.568 - 2.2975 + 2 * 3.234)`
`y_(4,p)=6.871`
So, the predicted value is `6.871`
Now, we will correct it by corrector method to get the final value
`y'_4=(x+y)/2=4.4355` (where `x=2,y=6.871`)
Milne's simpson corrector formula is
`y_(n+1,c) = y_(n-1) + h/3 (y'_(n-1) + 4y'_(n) + y'_(n+1))`
putting `n=3`, we get
`y_(4,c) = y_2 + h/3 (y'_2 + 4y'_3 + y'_4)`
`y_(4,c) = 3.595 + 0.5/3 * (2.2975 + 4 * 3.234 + 4.4355)`
`y_4=6.8732`
`:.y(2) = 6.8732` (Answer)
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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