Formula

4. Forth order R-K method `k_1=f(x_0,y_0,z_0)` `l_1=g(x_0,y_0,z_0)` `k_2=f(x_0+h/2,y_0+(hk_1)/2,z_0+(hl_1)/2)` `l_2=g(x_0+h/2,y_0+(hk_1)/2,z_0+(hl_1)/2)` `k_3=f(x_0+h/2,y_0+(hk_2)/2,z_0+(hl_2)/2)` `l_3=g(x_0+h/2,y_0+(hk_2)/2,z_0+(hl_2)/2)` `k_4=f(x_0+h,y_0+hk_3,z_0+hl_3)` `l_4=g(x_0+h,y_0+hk_3,z_0+hl_3)` `y_1=y_0+h/6(k_1+2k_2+2k_3+k_4)`

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Examples
Find y(0.1) for `y''=1+2xy-x^2z`, `x_0=0, y_0=1, z_0=0`, with step length 0.1 using Runge-Kutta 4 method (2nd order derivative)

Solution:
Given `y^('')=1+2xy-x^2z, y(0)=1, y'(0)=0, h=0.1, y(0.1)=?`

put `(dy)/(dx)=z` and differentiate w.r.t. x, we obtain `(d^2y)/(dx^2)=(dz)/(dx)`

We have system of equations
`(dy)/(dx)=z=f(x,y,z)`

`(dz)/(dx)=1+2xy-x^2z=g(x,y,z)`

Fourth order R-K method for second order differential equation
`k_1=f(x_0,y_0,z_0)=f(0,1,0)=0`

`l_1=g(x_0,y_0,z_0)=g(0,1,0)=1`

`k_2=f(x_0+h/2,y_0+(hk_1)/2,z_0+(hl_1)/2)=f(0.05,1,0.05)=0.05`

`l_2=g(x_0+h/2,y_0+(hk_1)/2,z_0+(hl_1)/2)=g(0.05,1,0.05)=1.0999`

`k_3=f(x_0+h/2,y_0+(hk_2)/2,z_0+(hl_2)/2)=f(0.05,1.0025,0.055)=0.055`

`l_3=g(x_0+h/2,y_0+(hk_2)/2,z_0+(hl_2)/2)=g(0.05,1.0025,0.055)=1.1001`

`k_4=f(x_0+h,y_0+hk_3,z_0+hl_3)=f(0.1,1.0055,0.11)=0.11`

`l_4=g(x_0+h,y_0+hk_3,z_0+hl_3)=g(0.1,1.0055,0.11)=1.2`

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

`y_1=1+0.1/6[0+2(0.05)+2(0.055)+(0.11)]`

`y_1=1.0053`

`:.y(0.1)=1.0053`

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`:.y(0.1)=1.0053`

`n`	`x_n`	`y_n`	`z_n`	`k_1`	`l_1`	`k_2`	`l_2`	`k_3`	`l_3`	`k_4`	`l_4`	`x_(n+1)`	`y_(n+1)`	`z_(n+1)`
0	0	1	0	0	1	0.05	1.0999	0.055	1.1001	0.11	1.2	0.1	1.0053

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