1. 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 3 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)`

Third order R-K method
`k_1=hf(x_0,y_0,z_0)=(0.1)*f(0,1,0)=(0.1)*(0)=0`

`l_1=hg(x_0,y_0,z_0)=(0.1)*g(0,1,0)=(0.1)*(1)=0.1`

`k_2=hf(x_0+h/2,y_0+k_1/2,z_0+l_1/2)=(0.1)*f(0.05,1,0.05)=(0.1)*(0.05)=0.005`

`l_2=hg(x_0+h/2,y_0+k_1/2,z_0+l_1/2)=(0.1)*g(0.05,1,0.05)=(0.1)*(1.0999)=0.11`

`k_3=hf(x_0+h,y_0+2k_2-k_1,z_0+2l_2-l_1)=(0.1)*f(0.1,1.01,0.12)=(0.1)*(0.12)=0.012`

`l_3=hg(x_0+h,y_0+2k_2-k_1,z_0+2l_2-l_1)=(0.1)*g(0.1,1.01,0.12)=(0.1)*(1.2008)=0.1201`

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

`y_1=1+1/6[0+4(0.005)+(0.012)]`

`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`	`x_(n+1)`	`y_(n+1)`	`z_(n+1)`
0	0	1	0	0	0.1	0.005	0.11	0.012	0.1201	0.1	1.0053