4. Formula-2 & Example-1
Formula
4. Forth order R-K method
`k_1=f(x_0,y_0)`
`k_2=f(x_0+h/2,y_0+(hk_1)/2)`
`k_3=f(x_0+h/2,y_0+(hk_2)/2)`
`k_4=f(x_0+h,y_0+hk_3)`
`y_1=y_0+h/6(k_1+2k_2+2k_3+k_4)`
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Examples
Find y(0.2) for `y'=(x-y)/2`, `x_0=0, y_0=1`, with step length 0.1 using Runge-Kutta 4 method (1st order derivative)
Solution: Given `y'=(x-y)/2, y(0)=1, h=0.1, y(0.2)=?`
Forth order R-K method `k_1=f(x_0,y_0)=f(0,1)=-0.5`
`k_2=f(x_0+h/2,y_0+(hk_1)/2)=f(0.05,0.975)=-0.4625`
`k_3=f(x_0+h/2,y_0+(hk_2)/2)=f(0.05,0.9769)=-0.4634`
`k_4=f(x_0+h,y_0+hk_3)=f(0.1,0.9537)=-0.4268`
`y_1=y_0+h/6(k_1+2k_2+2k_3+k_4)`
`y_1=1+0.1/6[-0.5+2(-0.4625)+2(-0.4634)+(-0.4268)]`
`y_1=0.9537`
`:.y(0.1)=0.9537`
Again taking `(x_1,y_1)` in place of `(x_0,y_0)` and repeat the process
`k_1=f(x_1,y_1)=f(0.1,0.9537)=-0.4268`
`k_2=f(x_1+h/2,y_1+(hk_1)/2)=f(0.15,0.9323)=-0.3912`
`k_3=f(x_1+h/2,y_1+(hk_2)/2)=f(0.15,0.9341)=-0.3921`
`k_4=f(x_1+h,y_1+hk_3)=f(0.2,0.9145)=-0.3572`
`y_2=y_1+h/6(k_1+2k_2+2k_3+k_4)`
`y_2=0.9537+0.1/6[-0.4268+2(-0.3912)+2(-0.3921)+(-0.3572)]`
`y_2=0.9145`
`:.y(0.2)=0.9145`
`:.y(0.2)=0.9145`
This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
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