Find y(0.2) for `y'=-y`, `x_0=0, y_0=1`, with step length 0.1 using Runge-Kutta 3 method (1st order derivative)
Solution:
Given `y'=-y, y(0)=1, h=0.1, y(0.2)=?`
Third 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`
`k_3=f(x_0+h,y_0+2hk_2-hk_1)=f(0.1,0.91)=-0.91`
`y_1=y_0+h/6(k_1+4k_2+k_3)`
`y_1=1+(0.1)/6[-1+4(-0.95)+(-0.91)]`
`y_1=0.90483`
`:.y(0.1)=0.90483`
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(0.1,0.90483)=-0.90483`
`k_2=f(x_1+h/2,y_1+(hk_1)/2)=f(0.15,0.85959)=-0.85959`
`k_3=f(x_1+h,y_1+2hk_2-hk_1)=f(0.2,0.8234)=-0.8234`
`y_2=y_1+h/6(k_1+4k_2+k_3)`
`y_2=0.90483+(0.1)/6[-0.90483+4(-0.85959)+(-0.8234)]`
`y_2=0.81872`
`:.y(0.2)=0.81872`
`:.y(0.2)=0.81872`
This material is intended as a summary. Use your textbook for detail explanation.
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