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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`

5. Example-2 : `y''=xz^2-y^2`





Find y(0.4) for `y''=xz^2-y^2`, `x_0=0, y_0=1, z_0=0`, with step length 0.2 using Runge-Kutta 2 method (second order differential equation)

Solution:
Given `y^('')=xz^2-y^2, y(0)=1, y'(0)=0, h=0.2, y(0.4)=?`

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)`

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(0.2,1,-0.2)`

`=-0.2`

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

`=g(0.2,1,-0.2)`

`=-0.992`

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

`=1-0.02`

`=0.98`

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

`=0-0.1992`

`=-0.1992`

`x_1=x_0+h=0+0.2=0.2`



for `n=1,x_1=0.2,y_1=0.98,z_1=-0.1992`

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

`=f(0.2,0.98,-0.1992)`

`=-0.1992`

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

`=g(0.2,0.98,-0.1992)`

`=-0.9525`

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

`=f(0.4,0.9402,-0.3897)`

`=-0.3897`

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

`=g(0.4,0.9402,-0.3897)`

`=-0.8232`

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

`=0.98-0.0589`

`=0.9211`

`x_2=x_1+h=0.2+0.2=0.4`

`:.y(0.4)=0.9211`

`n``x_n``y_n``z_n``k_1``l_1``k_2``l_2``x_(n+1)``y_(n+1)``z_(n+1)`
00100-1-0.2-0.9920.20.98-0.1992
10.20.98-0.1992-0.1992-0.9525-0.3897-0.82320.40.9211




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(0.1,1,-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`

`y_1=y_0+hk_2`

`=1-0.02`

`=0.98`

`z_1=z_0+hl_2`

`=0-0.1998`

`=-0.1998`

`x_1=x_0+h=0+0.2=0.2`



for `n=1,x_1=0.2,y_1=0.98,z_1=-0.1998`

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

`=f(0.2,0.98,-0.1998)`

`=-0.1998`

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

`=g(0.2,0.98,-0.1998)`

`=-0.9524`

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

`=f(0.3,0.96,-0.295)`

`=-0.295`

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

`=g(0.3,0.96,-0.295)`

`=-0.8955`

`y_2=y_1+hk_2`

`=0.98-0.059`

`=0.921`

`x_2=x_1+h=0.2+0.2=0.4`

`:.y(0.4)=0.921`

`n``x_n``y_n``z_n``k_1``l_1``k_2``l_2``x_(n+1)``y_(n+1)``z_(n+1)`
00100-1-0.1-0.9990.20.98-0.1998
10.20.98-0.1998-0.1998-0.9524-0.295-0.89550.40.921





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