Find y(0.4) for `y''=xz^2-y^2`, `x_0=0, y_0=1, z_0=0`, with step length 0.2 using Runge-Kutta 3 method (second order differential equation)

Solution:
Given `y^('')=xz^2-y^2, y(0)=1, y'(0)=0, h=0.2, y(0.4)=?`

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)=xz^2-y^2=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=1,z_0=0`

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

`=(0.2)*f(0,1,0)`

`=(0.2)*(0)`

`=0`

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

`=(0.2)*g(0,1,0)`

`=(0.2)*(-1)`

`=-0.2`

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

`=(0.2)*f(0.1,1,-0.1)`

`=(0.2)*(-0.1)`

`=-0.02`

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

`=(0.2)*g(0.1,1,-0.1)`

`=(0.2)*(-0.999)`

`=-0.1998`

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

`=(0.2)*f(0.2,0.96,-0.1996)`

`=(0.2)*(-0.1996)`

`=-0.0399`

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

`=(0.2)*g(0.2,0.96,-0.1996)`

`=(0.2)*(-0.9136)`

`=-0.1827`

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

`=1+1/6[0+4(-0.02)+(-0.0399)]`

`=0.98`

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

`=0+1/6[-0.2+4(-0.1998)+(-0.1827)]`

`=-0.197`

`x_1=x_0+h=0+0.2=0.2`

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for `n=1,x_1=0.2,y_1=0.98,z_1=-0.197`

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

`=(0.2)*f(0.2,0.98,-0.197)`

`=(0.2)*(-0.197)`

`=-0.0394`

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

`=(0.2)*g(0.2,0.98,-0.197)`

`=(0.2)*(-0.9527)`

`=-0.1905`

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

`=(0.2)*f(0.3,0.9603,-0.2923)`

`=(0.2)*(-0.2923)`

`=-0.0585`

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

`=(0.2)*g(0.3,0.9603,-0.2923)`

`=(0.2)*(-0.8966)`

`=-0.1793`

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

`=(0.2)*f(0.4,0.9025,-0.3651)`

`=(0.2)*(-0.3651)`

`=-0.073`

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

`=(0.2)*g(0.4,0.9025,-0.3651)`

`=(0.2)*(-0.7612)`

`=-0.1522`

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

`=0.98+1/6[-0.0394+4(-0.0585)+(-0.073)]`

`=0.9223`

`x_2=x_1+h=0.2+0.2=0.4`

`:.y(0.4)=0.9223`

`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	-0.2	-0.02	-0.1998	-0.0399	-0.1827	0.2	0.98	-0.197
1	0.2	0.98	-0.197	-0.0394	-0.1905	-0.0585	-0.1793	-0.073	-0.1522	0.4	0.9223

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