Find y(0.2) for `y''=-4z-4y`, `x_0=0, y_0=0, z_0=1`, with step length 0.1 using Runge-Kutta 3 method (second order differential equation)

Solution:
Given `y^('')=-4z-4y, y(0)=0, y'(0)=1, 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)=-4z-4y=g(x,y,z)`

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

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

`k_2=hf(x_n+h/2,y_n+k_1/2,z_n+l_1/2)`

`l_2=hg(x_n+h/2,y_n+k_1/2,z_n+l_1/2)`

`k_3=hf(x_n+h,y_n+2k_2-k_1,z_n+2l_2-l_1)`

`l_3=hg(x_n+h,y_n+2k_2-k_1,z_n+2l_2-l_1)`

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

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

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

`k_1=hf(x_0,y_0,z_0)`

`=(0.1)*f(0,0,1)`

`=(0.1)*(1)`

`=0.1`

`l_1=hg(x_0,y_0,z_0)`

`=(0.1)*g(0,0,1)`

`=(0.1)*(-4)`

`=-0.4`

`k_2=hf(x_0+h/2,y_0+k_1/2,z_0+l_1/2)`

`=(0.1)*f(0.05,0.05,0.8)`

`=(0.1)*(0.8)`

`=0.08`

`l_2=hg(x_0+h/2,y_0+k_1/2,z_0+l_1/2)`

`=(0.1)*g(0.05,0.05,0.8)`

`=(0.1)*(-3.4)`

`=-0.34`

`k_3=hf(x_0+h,y_0+2k_2-k_1,z_0+2l_2-l_1)`

`=(0.1)*f(0.1,0.06,0.72)`

`=(0.1)*(0.72)`

`=0.072`

`l_3=hg(x_0+h,y_0+2k_2-k_1,z_0+2l_2-l_1)`

`=(0.1)*g(0.1,0.06,0.72)`

`=(0.1)*(-3.12)`

`=-0.312`

Now,
`y_1=y_0+1/6(k_1+4k_2+k_3)`

`=0+1/6[0.1+4(0.08)+(0.072)]`

`=0.082`

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

`=1+1/6[-0.4+4(-0.34)+(-0.312)]`

`=0.6547`

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

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for `n=1,x_1=0.1,y_1=0.082,z_1=0.6547`

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

`=(0.1)*f(0.1,0.082,0.6547)`

`=(0.1)*(0.6547)`

`=0.0655`

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

`=(0.1)*g(0.1,0.082,0.6547)`

`=(0.1)*(-2.9467)`

`=-0.2947`

`k_2=hf(x_1+h/2,y_1+k_1/2,z_1+l_1/2)`

`=(0.1)*f(0.15,0.1147,0.5073)`

`=(0.1)*(0.5073)`

`=0.0507`

`l_2=hg(x_1+h/2,y_1+k_1/2,z_1+l_1/2)`

`=(0.1)*g(0.15,0.1147,0.5073)`

`=(0.1)*(-2.4883)`

`=-0.2488`

`k_3=hf(x_1+h,y_1+2k_2-k_1,z_1+2l_2-l_1)`

`=(0.1)*f(0.2,0.118,0.4517)`

`=(0.1)*(0.4517)`

`=0.0452`

`l_3=hg(x_1+h,y_1+2k_2-k_1,z_1+2l_2-l_1)`

`=(0.1)*g(0.2,0.118,0.4517)`

`=(0.1)*(-2.2787)`

`=-0.2279`

Now,
`y_2=y_1+1/6(k_1+4k_2+k_3)`

`=0.082+1/6[0.0655+4(0.0507)+(0.0452)]`

`=0.1343`

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

`:.y(0.2)=0.1343`

`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	0	1	0.1	-0.4	0.08	-0.34	0.072	-0.312	0.1	0.082	0.6547
1	0.1	0.082	0.6547	0.0655	-0.2947	0.0507	-0.2488	0.0452	-0.2279	0.2	0.1343

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