Formula

Third order R-K method `k_1=hf(x_n,y_n)` `k_2=hf(x_n+h/2,y_n+k_1/2)` `k_3=hf(x_n+h,y_n+2k_2-k_1)` `y_(n+1)=y_n+1/6(k_1+4k_2+k_3)`

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Examples
1. Find y(0.2) for `y'=(x-y)/2`, `x_0=0, y_0=1`, with step length 0.1 using Runge-Kutta 3 method (first order differential equation)

Solution:
Given `y'=(x-y)/(2), y(0)=1, h=0.1, y(0.2)=?`

Third order Runge-Kutta (RK3) method formula
`k_1=hf(x_n,y_n)`

`k_2=hf(x_n+h/2,y_n+k_1/2)`

`k_3=hf(x_n+h,y_n+2k_2-k_1)`

`y_(n+1)=y_n+1/6(k_1+4k_2+k_3)`

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for `n=0,x_0=0,y_0=1`

`k_1=hf(x_0,y_0)`

`=(0.1)f(0,1)`

`=(0.1)*(-0.5)`

`=-0.05`

`k_2=hf(x_0+h/2,y_0+k_1/2)`

`=(0.1)f(0.05,0.975)`

`=(0.1)*(-0.4625)`

`=-0.0462`

`k_3=hf(x_0+h,y_0+2k_2-k_1)`

`=(0.1)f(0.1,0.9575)`

`=(0.1)*(-0.4288)`

`=-0.0429`

`y_1=y_0+1/6(k_1+4k_2+k_3)`

`=1+1/6[-0.05+4(-0.0462)+(-0.0429)]`

`=0.9537`

`x_1=x_0+h=0+0.1=0.1`

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for `n=1,x_1=0.1,y_1=0.9537`

`k_1=hf(x_1,y_1)`

`=(0.1)f(0.1,0.9537)`

`=(0.1)*(-0.4268)`

`=-0.0427`

`k_2=hf(x_1+h/2,y_1+k_1/2)`

`=(0.1)f(0.15,0.9323)`

`=(0.1)*(-0.3912)`

`=-0.0391`

`k_3=hf(x_1+h,y_1+2k_2-k_1)`

`=(0.1)f(0.2,0.9181)`

`=(0.1)*(-0.3591)`

`=-0.0359`

`y_2=y_1+1/6(k_1+4k_2+k_3)`

`=0.9537+1/6[-0.0427+4(-0.0391)+(-0.0359)]`

`=0.9145`

`x_2=x_1+h=0.1+0.1=0.2`

`:.y(0.2)=0.9145`

`n`	`x_n`	`y_n`	`k_1`	`k_2`	`k_3`	`x_(n+1)`	`y_(n+1)`
0	0	1	-0.05	-0.0462	-0.0429	0.1	0.9537
1	0.1	0.9537	-0.0427	-0.0391	-0.0359	0.2	0.9145

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