Find y(0.2) for `y''=xz^2-y^2`, `x_0=0, y_0=1, z_0=0`, with step length 0.2 using Runge-Kutta 4 method (2nd order derivative)

Solution:
Given `y^('')=xz^2-y^2, y(0)=1, y'(0)=0, h=0.2, 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)=xz^2-y^2=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(0.1,1,-0.1)=-0.1`

`l_2=g(x_0+h/2,y_0+(hk_1)/2,z_0+(hl_1)/2)=g(0.1,1,-0.1)=-0.999`

`k_3=f(x_0+h/2,y_0+(hk_2)/2,z_0+(hl_2)/2)=f(0.1,0.99,-0.0999)=-0.0999`

`l_3=g(x_0+h/2,y_0+(hk_2)/2,z_0+(hl_2)/2)=g(0.1,0.99,-0.0999)=-0.9791`

`k_4=f(x_0+h,y_0+hk_3,z_0+hl_3)=f(0.2,0.98,-0.1958)=-0.1958`

`l_4=g(x_0+h,y_0+hk_3,z_0+hl_3)=g(0.2,0.98,-0.1958)=-0.9528`

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

`y_1=1+0.2/6[0+2(-0.1)+2(-0.0999)+(-0.1958)]`

`y_1=0.9801`

`:.y(0.2)=0.9801`

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`:.y(0.2)=0.9801`

`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	-0.1	-0.999	-0.0999	-0.9791	-0.1958	-0.9528	0.2	0.9801

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