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=hf(x_0,y_0)=(0.1)f(0,1)=(0.1)*(-1)=-0.1`
`k_2=hf(x_0+h/2,y_0+k_1/2)=(0.1)f(0.05,0.95)=(0.1)*(-0.95)=-0.095`
`k_3=hf(x_0+h,y_0+2k_2-k_1)=(0.1)f(0.1,0.91)=(0.1)*(-0.91)=-0.091`
`y_1=y_0+1/6(k_1+4k_2+k_3)`
`y_1=1+1/6[-0.1+4(-0.095)+(-0.091)]`
`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=hf(x_1,y_1)=(0.1)f(0.1,0.90483)=(0.1)*(-0.90483)=-0.09048`
`k_2=hf(x_1+h/2,y_1+k_1/2)=(0.1)f(0.15,0.85959)=(0.1)*(-0.85959)=-0.08596`
`k_3=hf(x_1+h,y_1+2k_2-k_1)=(0.1)f(0.2,0.8234)=(0.1)*(-0.8234)=-0.08234`
`y_2=y_1+1/6(k_1+4k_2+k_3)`
`y_2=0.90483+1/6[-0.09048+4(-0.08596)+(-0.08234)]`
`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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