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11. Runge-Kutta 4 method (second order differential equation) example ( Enter your problem )
  1. Formula & Example-1 : `y''=1+2xy-x^2z`
  2. Example-2 : `y''=xz^2-y^2`
  3. Example-3 : `y''=-4z-4y`
  4. Formula-2 & Example-1 : `y''=1+2xy-x^2z`
  5. Example-2 : `y''=xz^2-y^2`
  6. Example-3 : `y''=-4z-4y`

1. Formula & Example-1 : `y''=1+2xy-x^2z`





Formula
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)`

Examples
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)`
001000.10.0050.110.00550.110.0110.120.11.00530.11
10.11.00530.110.0110.120.0170.12990.01750.130.0240.140.21.0227





This material is intended as a summary. Use your textbook for detail explanation.
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