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4. Runge-Kutta 4 method (2nd order derivative) example ( Enter your problem )
  1. Formula-1 & Example-1
  2. Example-2
  3. Example-3
  4. Formula-2 & Example-1
  5. Example-2
  6. Example-3
Other related methods
  1. Euler method (1st order derivative)
  2. Runge-Kutta 2 method (1st order derivative)
  3. Runge-Kutta 3 method (1st order derivative)
  4. Runge-Kutta 4 method (1st order derivative)
  5. Improved Euler method (1st order derivative)
  6. Modified Euler method (1st order derivative)
  7. Taylor Series method (1st order derivative)
  8. Euler method (2nd order derivative)
  9. Runge-Kutta 2 method (2nd order derivative)
  10. Runge-Kutta 3 method (2nd order derivative)
  11. Runge-Kutta 4 method (2nd order derivative)

10. Runge-Kutta 3 method (2nd order derivative)
(Previous method)
2. Example-2
(Next example)

1. Formula-1 & Example-1





Formula
4. Forth order R-K method
`k_1=hf(x_0,y_0,z_0)`
`l_1=hg(x_0,y_0,z_0)`
`k_2=hf(x_0+h/2,y_0+k_1/2,z_0+l_1/2)`
`l_2=hg(x_0+h/2,y_0+k_1/2,z_0+l_1/2)`
`k_3=hf(x_0+h/2,y_0+k_2/2,z_0+l_2/2)`
`l_3=hg(x_0+h/2,y_0+k_2/2,z_0+l_2/2)`
`k_4=hf(x_0+h,y_0+k_3,z_0+l_3)`
`l_4=hg(x_0+h,y_0+k_3,z_0+l_3)`
`y_1=y_0+1/6(k_1+2k_2+2k_3+k_4)`

Examples
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 4 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)`

Forth order R-K method for second order differential equation
`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)*(1.09988)=0.10999`

`k_3=hf(x_0+h/2,y_0+k_2/2,z_0+l_2/2)=(0.1)*f(0.05,1.0025,0.05499)=(0.1)*(0.05499)=0.0055`

`l_3=hg(x_0+h/2,y_0+k_2/2,z_0+l_2/2)=(0.1)*g(0.05,1.0025,0.05499)=(0.1)*(1.10011)=0.11001`

`k_4=hf(x_0+h,y_0+k_3,z_0+l_3)=(0.1)*f(0.1,1.0055,0.11001)=(0.1)*(0.11001)=0.011`

`l_4=hg(x_0+h,y_0+k_3,z_0+l_3)=(0.1)*g(0.1,1.0055,0.11001)=(0.1)*(1.2)=0.12`

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

`y_1=1+1/6[0+2(0.005)+2(0.0055)+(0.011)]`

`y_1=1.00533`

`:.y(0.1)=1.00533`



`:.y(0.1)=1.00533`


This material is intended as a summary. Use your textbook for detail explanation.
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10. Runge-Kutta 3 method (2nd order derivative)
(Previous method)
2. Example-2
(Next example)





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