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9. Runge-Kutta 2 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`

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





Formula
Second order Runge-Kutta (RK2) method for second order differential equation formula
Method-1 :
`k_1=f(x_n,y_n,z_n)`
`l_1=g(x_n,y_n,z_n)`
`k_2=f(x_n+h,y_n+hk_1,z_n+hl_1)`
`l_2=g(x_n+h,y_n+hk_1,z_n+hl_1)`
`y_(n+1)=y_n+(h(k_1+k_2))/2`

Method-2 :
`k_1=f(x_n,y_n,z_n)`
`l_1=g(x_n,y_n,z_n)`
`k_2=f(x_n+h/2,y_n+(hk_1)/2,z_n+(hl_1)/2)`
`l_2=g(x_n+h/2,y_n+(hk_1)/2,z_n+(hl_1)/2)`
`y_(n+1)=y_n+hk_2`

Examples
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 2 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)`

Method-1 : Using formula `k_2=f(x_0+h,y_0+hk_1,z_0+hl_1)`

Second order Runge-Kutta (RK2) method for second order differential equation formula
`k_1=f(x_n,y_n,z_n)`

`l_1=g(x_n,y_n,z_n)`

`k_2=f(x_n+h,y_n+hk_1,z_n+hl_1)`

`l_2=g(x_n+h,y_n+hk_1,z_n+hl_1)`

`y_(n+1)=y_n+(h(k_1+k_2))/2`

`z_(n+1)=z_n+(h(l_1+l_2))/2`



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

`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,y_0+hk_1,z_0+hl_1)`

`=f(0.1,1,0.1)`

`=0.1`

`l_2=g(x_0+h,y_0+hk_1,z_0+hl_1)`

`=g(0.1,1,0.1)`

`=1.199`

`y_1=y_0+(h(k_1+k_2))/2`

`=1+0.005`

`=1.005`

`z_1=z_0+(h(l_1+l_2))/2`

`=0+0.11`

`=0.11`

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



for `n=1,x_1=0.1,y_1=1.005,z_1=0.11`

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

`=f(0.1,1.005,0.11)`

`=0.11`

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

`=g(0.1,1.005,0.11)`

`=1.1999`

`k_2=f(x_1+h,y_1+hk_1,z_1+hl_1)`

`=f(0.2,1.016,0.2299)`

`=0.2299`

`l_2=g(x_1+h,y_1+hk_1,z_1+hl_1)`

`=g(0.2,1.016,0.2299)`

`=1.3972`

`y_2=y_1+(h(k_1+k_2))/2`

`=1.005+0.017`

`=1.022`

`x_2=x_1+h=0.1+0.1=0.2`

`:.y(0.2)=1.022`

`n``x_n``y_n``z_n``k_1``l_1``k_2``l_2``x_(n+1)``y_(n+1)``z_(n+1)`
0010010.11.1990.11.0050.11
10.11.0050.110.111.19990.22991.39720.21.022




Method-2 : Using formula `k_2=f(x_0+h/2,y_0+(hk_1)/2,z_0+(hl_1)/2)`

Second order Runge-Kutta (RK2) method for second order differential equation formula
`k_1=f(x_n,y_n,z_n)`

`l_1=g(x_n,y_n,z_n)`

`k_2=f(x_n+h/2,y_n+(hk_1)/2,z_n+(hl_1)/2)`

`l_2=g(x_n+h/2,y_n+(hk_1)/2,z_n+(hl_1)/2)`

`y_(n+1)=y_n+hk_2`

`z_(n+1)=z_n+hl_2`



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

`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(0.05,1,0.05)`

`=0.05`

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

`=g(0.05,1,0.05)`

`=1.0999`

`y_1=y_0+hk_2`

`=1+0.005`

`=1.005`

`z_1=z_0+hl_2`

`=0+0.11`

`=0.11`

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



for `n=1,x_1=0.1,y_1=1.005,z_1=0.11`

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

`=f(0.1,1.005,0.11)`

`=0.11`

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

`=g(0.1,1.005,0.11)`

`=1.1999`

`k_2=f(x_1+h/2,y_1+(hk_1)/2,z_1+(hl_1)/2)`

`=f(0.15,1.0105,0.17)`

`=0.17`

`l_2=g(x_1+h/2,y_1+(hk_1)/2,z_1+(hl_1)/2)`

`=g(0.15,1.0105,0.17)`

`=1.2993`

`y_2=y_1+hk_2`

`=1.005+0.017`

`=1.022`

`x_2=x_1+h=0.1+0.1=0.2`

`:.y(0.2)=1.022`

`n``x_n``y_n``z_n``k_1``l_1``k_2``l_2``x_(n+1)``y_(n+1)``z_(n+1)`
0010010.051.09990.11.0050.11
10.11.0050.110.111.19990.171.29930.21.022





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