1. Formula-1 & Example-1
Formula
4. Forth order R-K method
`k_1=hf(x_0,y_0)`
`k_2=hf(x_0+h/2,y_0+k_1/2)`
`k_3=hf(x_0+h/2,y_0+k_2/2)`
`k_4=hf(x_0+h,y_0+k_3)`
`y_1=y_0+1/6(k_1+2k_2+2k_3+k_4)`
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Examples
1. 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=hf(x_0,y_0)=(0.1)f(0,1)=(0.1)*(-0.5)=-0.05`
`k_2=hf(x_0+h/2,y_0+k_1/2)=(0.1)f(0.05,0.975)=(0.1)*(-0.4625)=-0.04625`
`k_3=hf(x_0+h/2,y_0+k_2/2)=(0.1)f(0.05,0.97688)=(0.1)*(-0.46344)=-0.04634`
`k_4=hf(x_0+h,y_0+k_3)=(0.1)f(0.1,0.95366)=(0.1)*(-0.42683)=-0.04268`
`y_1=y_0+1/6(k_1+2k_2+2k_3+k_4)`
`y_1=1+1/6[-0.05+2(-0.04625)+2(-0.04634)+(-0.04268)]`
`y_1=0.95369`
`:.y(0.1)=0.95369`
Again taking `(x_1,y_1)` in place of `(x_0,y_0)` and repeat the process
`k_1=hf(x_1,y_1)=(0.1)f(0.1,0.95369)=(0.1)*(-0.42684)=-0.04268`
`k_2=hf(x_1+h/2,y_1+k_1/2)=(0.1)f(0.15,0.93235)=(0.1)*(-0.39117)=-0.03912`
`k_3=hf(x_1+h/2,y_1+k_2/2)=(0.1)f(0.15,0.93413)=(0.1)*(-0.39206)=-0.03921`
`k_4=hf(x_1+h,y_1+k_3)=(0.1)f(0.2,0.91448)=(0.1)*(-0.35724)=-0.03572`
`y_2=y_1+1/6(k_1+2k_2+2k_3+k_4)`
`y_2=0.95369+1/6[-0.04268+2(-0.03912)+2(-0.03921)+(-0.03572)]`
`y_2=0.91451`
`:.y(0.2)=0.91451`
`:.y(0.2)=0.91451`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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