Formula

Third 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,y_n+2hk_2-hk_1)` `y_(n+1)=y_n+h/6(k_1+4k_2+k_3)`

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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 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=f(x_n,y_n)`

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

`k_3=f(x_n+h,y_n+2hk_2-hk_1)`

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

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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,y_0+2hk_2-hk_1)`

`=f(0.1,0.9575)`

`=-0.4288`

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

`=1+(0.1)/6[-0.5+4(-0.4625)+(-0.4288)]`

`=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,y_1+2hk_2-hk_1)`

`=f(0.2,0.9181)`

`=-0.3591`

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

`=0.9537+(0.1)/6[-0.4268+4(-0.3912)+(-0.3591)]`

`=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	-0.5	-0.4625	-0.4288	0.1	0.9537
1	0.1	0.9537	-0.4268	-0.3912	-0.3591	0.2	0.9145

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