1. Formula & Example-1 `y'=(x+y)/2` (table data)
Formula
Adam's Bashforth Predictor formula is
`y_(n+1,p) = y_n + h/24 (55y'_(n) - 59y'_(n-1) + 37y'_(n-2) - 9y'_(n-3))`
putting `n=3`, we get
`y_(4,p)=y_3 + h/24 (55y'_(3) - 59y'_2 + 37y'_1 - 9y'_0)`
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Adam's Bashforth Corrector formula is
`y_(n+1,c) = y_n + h/24 (9y'_(n+1) + 19y'_(n) - 5y'_(n-1) + y'_(n-2))`
putting `n=3`, we get
`y_(4,c) = y_3 + h/24 (9y'_4 + 19y'_3 - 5y'_2 + y'_1)`
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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 Adams bashforth predictor method
Solution:
`y'=(x+y)/2`
Adam's Bashforth Predictor formula is
`y_(n+1,p) = y_n + h/24 (55y'_(n) - 59y'_(n-1) + 37y'_(n-2) - 9y'_(n-3))`
putting `n=3`, we get
`y_(4,p)=y_3 + h/24 (55y'_(3) - 59y'_2 + 37y'_1 - 9y'_0) ->(2)`
We have given that
`x_0=0,x_1=0.5,x_2=1,x_3=1.5`
`y_0=2,y_1=2.636,y_2=3.595,y_3=4.968`
`y'=(x+y)/2`
`y'_0=(x+y)/2=1` (where `x=0,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_3 + h/24 (55y'_(3) - 59y'_2 + 37y'_1 - 9y'_0)`
`y_(4,p)=4.968 + 0.5/24 * (55 * 3.234 - 59 * 2.2975 + 37 * 1.568 - 9 * 1)`
`y_(4,p)=4.968 + 0.5/24 * (177.87 - 135.5525 + 58.016 - 9)`
`y_(4,p)=4.968 + 0.5/24 * (91.3335)`
`y_(4,p)=4.968 +1.9028`
`y_(4,p)=6.8708`
So, the predicted value is `6.8708`
Now, we will correct it by corrector method to get the final value
`y'_4=(x+y)/2=4.4354` (where `x=2,y=6.8708`)
Adam's Bashforth Corrector formula is
`y_(n+1,c) = y_n + h/24 (9y'_(n+1) + 19y'_(n) - 5y'_(n-1) + y'_(n-2))`
putting `n=3`, we get
`y_(4,c) = y_3 + h/24 (9y'_4 + 19y'_3 - 5y'_2 + y'_1)`
`y_(4,c) = 4.968 + 0.5/24 * (9 * 4.4354 + 19 * 3.234 - 5 * 2.2975 + 1.568)`
`y_(4,c) = 4.968 + 0.5/24 * (39.9185 + 61.446 - 11.4875 + 1.568)`
`y_(4,c) = 4.968 + 0.5/24 * (91.445)`
`y_(4,c) = 4.968 +1.9051`
`y_(4,c)=6.8731`
`:.y(2) = 6.8731`
This material is intended as a summary. Use your textbook for detail explanation.
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