4. Formula-2 & Example-1
Formula
3. Third order R-K method
`k_1=f(x_0,y_0,z_0)`
`l_1=g(x_0,y_0,z_0)`
`k_2=f(x_0+h/2,y_0+(hk_1)/2,z_0+(hl_1)/2)`
`l_2=g(x_0+h/2,y_0+(hk_1)/2,z_0+(hl_1)/2)`
`k_3=f(x_0+h,y_0+2hk_2-hk_1,z_0+2hl_2-hl_1)`
`l_3=g(x_0+h,y_0+2hk_2-hk_1,z_0+2hl_2-hl_1)`
`y_1=y_0+h/6(k_1+4k_2+k_3)`
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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 3 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)`
Third order R-K method
`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)=1.09988`
`k_3=f(x_0+h,y_0+2hk_2-hk_1,z_0+2hl_2-hl_1)=f(0.1,1.01,0.11998)=0.11998`
`l_3=g(x_0+h,y_0+2hk_2-hk_1,z_0+2hl_2-hl_1)=g(0.1,1.01,0.11998)=1.2008`
Now,
`y_1=y_0+h/6(k_1+4k_2+k_3)`
`y_1=1+0.1/6[0+4(0.05)+(0.11998)]`
`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.
Any bug, improvement, feedback then
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