6. Example-3
Find y(0.2) for `y'=-y`, `x_0=0, y_0=1`, with step length 0.1 using Runge-Kutta 2 method (1st order derivative)
Solution:
Given `y'=-y, y(0)=1, h=0.1, y(0.2)=?`
Method-1 : Using formula `k_2=f(x_0+h,y_0+hk_1)`
Second order R-K method
`k_1=f(x_0,y_0)=f(0,1)=-1`
`k_2=f(x_0+h,y_0+hk_1)=f(0.1,0.9)=-0.9`
`y_1=y_0+h/2(k_1+k_2)=1-0.095=0.905`
`:.y(0.1)=0.905`
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(0.1,0.905)=-0.905`
`k_2=f(x_1+h,y_1+hk_1)=f(0.2,0.8145)=-0.8145`
`y_2=y_1+h/2(k_1+k_2)=0.905-0.08598=0.81902`
`:.y(0.2)=0.81902`
`:.y(0.2)=0.81902`
Method-2 : Using formula `k_2=f(x_0+h/2,y_0+(hk_1)/2)`
Second order R-K method
`k_1=f(x_0,y_0)=f(0,1)=-1`
`k_2=f(x_0+h/2,y_0+(hk_1)/2)=f(0.05,0.95)=-0.95`
`y_1=y_0+hk_2=1-0.095=0.905`
`:.y(0.1)=0.905`
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(0.1,0.905)=-0.905`
`k_2=f(x_1+h/2,y_1+(hk_1)/2)=f(0.15,0.85975)=-0.85975`
`y_2=y_1+hk_2=0.905-0.08598=0.81902`
`:.y(0.2)=0.81902`
`:.y(0.2)=0.81902`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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