Formula

Forth order R-K method `k_1=f(x_n,y_n)` `k_2=f(x_n+h/2,y_n+(hk_1)/2)` `k_3=f(x_n+h/2,y_n+(hk_2)/2)` `k_4=f(x_n+h,y_n+hk_3)` `y_(n+1)=y_n+h/6(k_1+2k_2+2k_3+k_4)`

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Examples
Find y(0.2) for `y'=(x-y)/2`, `x_0=0, y_0=1`, with step length 0.1 using Runge-Kutta 4 method (first order differential equation)

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

Fourth order Runge-Kutta (RK4) method formula
`k_1=f(x_n,y_n)`

`k_2=f(x_n+h/2,y_n+(hk_1)/2)`

`k_3=f(x_n+h/2,y_n+(hk_2)/2)`

`k_4=f(x_n+h,y_n+hk_3)`

`y_(n+1)=y_n+h/6(k_1+2k_2+2k_3+k_4)`

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

`k_1=f(x_0,y_0)`

`=f(0,1)`

`=-0.5`

`k_2=f(x_0+h/2,y_0+(hk_1)/2)`

`=f(0.05,0.975)`

`=-0.4625`

`k_3=f(x_0+h/2,y_0+(hk_2)/2)`

`=f(0.05,0.9769)`

`=-0.4634`

`k_4=f(x_0+h,y_0+hk_3)`

`=f(0.1,0.9537)`

`=-0.4268`

`y_1=y_0+h/6(k_1+2k_2+2k_3+k_4)`

`=1+0.1/6[-0.5+2(-0.4625)+2(-0.4634)+(-0.4268)]`

`=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=f(x_1,y_1)`

`=f(0.1,0.9537)`

`=-0.4268`

`k_2=f(x_1+h/2,y_1+(hk_1)/2)`

`=f(0.15,0.9323)`

`=-0.3912`

`k_3=f(x_1+h/2,y_1+(hk_2)/2)`

`=f(0.15,0.9341)`

`=-0.3921`

`k_4=f(x_1+h,y_1+hk_3)`

`=f(0.2,0.9145)`

`=-0.3572`

`y_2=y_1+h/6(k_1+2k_2+2k_3+k_4)`

`=0.9537+0.1/6[-0.4268+2(-0.3912)+2(-0.3921)+(-0.3572)]`

`=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`	`k_4`	`x_(n+1)`	`y_(n+1)`
0	0	1	-0.5	-0.4625	-0.4634	-0.4268	0.1	0.9537
1	0.1	0.9537	-0.4268	-0.3912	-0.3921	-0.3572	0.2	0.9145

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