Formula

Third order Runge-Kutta (RK3) method for second order differential equation formula `k_1=f(x_n,y_n,z_n)` `l_1=g(x_n,y_n,z_n)` `k_2=f(x_n+h/2,y_n+(hk_1)/2,z_n+(hl_1)/2)` `l_2=g(x_n+h/2,y_n+(hk_1)/2,z_n+(hl_1)/2)` `k_3=f(x_n+h,y_n+2hk_2-hk_1,z_n+2hl_2-hl_1)` `l_3=g(x_n+h,y_n+2hk_2-hk_1,z_n+2hl_2-hl_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''=1+2xy-x^2z`, `x_0=0, y_0=1, z_0=0`, with step length 0.1 using Runge-Kutta 3 method (second order differential equation)

Solution:
Given `y^('')=1+2xy-x^2z, y(0)=1, y'(0)=0, h=0.1, y(0.2)=?`

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)`

Third order Runge-Kutta (RK3) method for second order differential equation formula
`k_1=f(x_n,y_n,z_n)`

`l_1=g(x_n,y_n,z_n)`

`k_2=f(x_n+h/2,y_n+(hk_1)/2,z_n+(hl_1)/2)`

`l_2=g(x_n+h/2,y_n+(hk_1)/2,z_n+(hl_1)/2)`

`k_3=f(x_n+h,y_n+2hk_2-hk_1,z_n+2hl_2-hl_1)`

`l_3=g(x_n+h,y_n+2hk_2-hk_1,z_n+2hl_2-hl_1)`

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

`z_(n+1)=z_n+h/6(l_1+4l_2+l_3)`

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for `n=0,x_0=0,y_0=1,z_0=0`

`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,y_0+2hk_2-hk_1,z_0+2hl_2-hl_1)`

`=f(0.1,1.01,0.12)`

`=0.12`

`l_3=g(x_0+h,y_0+2hk_2-hk_1,z_0+2hl_2-hl_1)`

`=g(0.1,1.01,0.12)`

`=1.2008`

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

`=1+0.1/6[0+4(0.05)+(0.12)]`

`=1.0053`

`z_1=z_0+1/6(l_1+4l_2+l_3)`

`=0+1/6[1+4(1.0999)+(1.2008)]`

`=0.11`

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

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for `n=1,x_1=0.1,y_1=1.0053,z_1=0.11`

`k_1=f(x_1,y_1,z_1)`

`=f(0.1,1.0053,0.11)`

`=0.11`

`l_1=g(x_1,y_1,z_1)`

`=g(0.1,1.0053,0.11)`

`=1.2`

`k_2=f(x_1+h/2,y_1+(hk_1)/2,z_1+(hl_1)/2)`

`=f(0.15,1.0108,0.17)`

`=0.17`

`l_2=g(x_1+h/2,y_1+(hk_1)/2,z_1+(hl_1)/2)`

`=g(0.15,1.0108,0.17)`

`=1.2994`

`k_3=f(x_1+h,y_1+2hk_2-hk_1,z_1+2hl_2-hl_1)`

`=f(0.2,1.0283,0.2499)`

`=0.2499`

`l_3=g(x_1+h,y_1+2hk_2-hk_1,z_1+2hl_2-hl_1)`

`=g(0.2,1.0283,0.2499)`

`=1.4013`

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

`=1.0053+0.1/6[0.11+4(0.17)+(0.2499)]`

`=1.0227`

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

`:.y(0.2)=1.0227`

`n`	`x_n`	`y_n`	`z_n`	`k_1`	`l_1`	`k_2`	`l_2`	`k_3`	`l_3`	`x_(n+1)`	`y_(n+1)`	`z_(n+1)`
0	0	1	0	0	1	0.05	1.0999	0.12	1.2008	0.1	1.0053	0.11
1	0.1	1.0053	0.11	0.11	1.2	0.17	1.2994	0.2499	1.4013	0.2	1.0227

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