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)