1. 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 4 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)`

Fourth order Runge-Kutta (RK4) 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/2,y_n+k_2/2,z_n+l_2/2)`

`l_3=hg(x_n+h/2,y_n+k_2/2,z_n+l_2/2)`

`k_4=hf(x_n+h,y_n+k_3,z_n+l_3)`

`l_4=hg(x_n+h,y_n+k_3,z_n+l_3)`

`y_(n+1)=y_n+1/6(k_1+2k_2+2k_3+k_4)`

`z_(n+1)=z_n+1/6(l_1+2l_2+2l_3+l_4)`

---

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

`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/2,y_0+k_2/2,z_0+l_2/2)`

`=(0.1)*f(0.05,1.0025,0.055)`

`=(0.1)*(0.055)`

`=0.0055`

`l_3=hg(x_0+h/2,y_0+k_2/2,z_0+l_2/2)`

`=(0.1)*g(0.05,1.0025,0.055)`

`=(0.1)*(1.1001)`

`=0.11`

`k_4=hf(x_0+h,y_0+k_3,z_0+l_3)`

`=(0.1)*f(0.1,1.0055,0.11)`

`=(0.1)*(0.11)`

`=0.011`

`l_4=hg(x_0+h,y_0+k_3,z_0+l_3)`

`=(0.1)*g(0.1,1.0055,0.11)`

`=(0.1)*(1.2)`

`=0.12`

Now,
`y_1=y_0+1/6(k_1+2k_2+2k_3+k_4)`

`=1+1/6[0+2(0.005)+2(0.0055)+(0.011)]`

`=1.0053`

`z_1=z_0+1/6(l_1+2l_2+2l_3+l_4)`

`=0+1/6[0.1+2(0.11)+2(0.11)+(0.12)]`

`=0.11`

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

---

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

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

`=(0.1)*f(0.1,1.0053,0.11)`

`=(0.1)*(0.11)`

`=0.011`

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

`=(0.1)*g(0.1,1.0053,0.11)`

`=(0.1)*(1.2)`

`=0.12`

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

`=(0.1)*f(0.15,1.0108,0.17)`

`=(0.1)*(0.17)`

`=0.017`

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

`=(0.1)*g(0.15,1.0108,0.17)`

`=(0.1)*(1.2994)`

`=0.1299`

`k_3=hf(x_1+h/2,y_1+k_2/2,z_1+l_2/2)`

`=(0.1)*f(0.15,1.0138,0.175)`

`=(0.1)*(0.175)`

`=0.0175`

`l_3=hg(x_1+h/2,y_1+k_2/2,z_1+l_2/2)`

`=(0.1)*g(0.15,1.0138,0.175)`

`=(0.1)*(1.3002)`

`=0.13`

`k_4=hf(x_1+h,y_1+k_3,z_1+l_3)`

`=(0.1)*f(0.2,1.0228,0.24)`

`=(0.1)*(0.24)`

`=0.024`

`l_4=hg(x_1+h,y_1+k_3,z_1+l_3)`

`=(0.1)*g(0.2,1.0228,0.24)`

`=(0.1)*(1.3995)`

`=0.14`

Now,
`y_2=y_1+1/6(k_1+2k_2+2k_3+k_4)`

`=1.0053+1/6[0.011+2(0.017)+2(0.0175)+(0.024)]`

`=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`	`k_4`	`l_4`	`x_(n+1)`	`y_(n+1)`	`z_(n+1)`
0	0	1	0	0	0.1	0.005	0.11	0.0055	0.11	0.011	0.12	0.1	1.0053	0.11
1	0.1	1.0053	0.11	0.011	0.12	0.017	0.1299	0.0175	0.13	0.024	0.14	0.2	1.0227