Formula

4. Forth order R-K method `k_1=hf(x_0,y_0)` `k_2=hf(x_0+h/2,y_0+k_1/2)` `k_3=hf(x_0+h/2,y_0+k_2/2)` `k_4=hf(x_0+h,y_0+k_3)` `y_1=y_0+1/6(k_1+2k_2+2k_3+k_4)`

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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 4 method (1st order derivative)

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

Fourth order R-K method
`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/2,y_0+k_2/2)=(0.1)f(0.05,0.9769)=(0.1)*(-0.4634)=-0.0463`

`k_4=hf(x_0+h,y_0+k_3)=(0.1)f(0.1,0.9537)=(0.1)*(-0.4268)=-0.0427`

`y_1=y_0+1/6(k_1+2k_2+2k_3+k_4)`

`y_1=1+1/6[-0.05+2(-0.0462)+2(-0.0463)+(-0.0427)]`

`y_1=0.9537`

`:.y(0.1)=0.9537`

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Again taking `(x_1,y_1)` in place of `(x_0,y_0)` and repeat the process

`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/2,y_1+k_2/2)=(0.1)f(0.15,0.9341)=(0.1)*(-0.3921)=-0.0392`

`k_4=hf(x_1+h,y_1+k_3)=(0.1)f(0.2,0.9145)=(0.1)*(-0.3572)=-0.0357`

`y_2=y_1+1/6(k_1+2k_2+2k_3+k_4)`

`y_2=0.9537+1/6[-0.0427+2(-0.0391)+2(-0.0392)+(-0.0357)]`

`y_2=0.9145`

`:.y(0.2)=0.9145`

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`:.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	-0.05	-0.0462	-0.0463	-0.0427	0.1	0.9537
1	0.1	0.9537	-0.0427	-0.0391	-0.0392	-0.0357	0.2	0.9145

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