3. 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=hf(x_0+h,y_0+k_1)`
Second 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,y_0+k_1)=(0.1)f(0.1,0.9)=(0.1)*(-0.9)=-0.09`
`y_1=y_0+(k_1+k_2)/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=hf(x_1,y_1)=(0.1)f(0.1,0.905)=(0.1)*(-0.905)=-0.0905`
`k_2=hf(x_1+h,y_1+k_1)=(0.1)f(0.2,0.8145)=(0.1)*(-0.8145)=-0.08145`
`y_2=y_1+(k_1+k_2)/2=0.905-0.08598=0.81902`
`:.y(0.2)=0.81902`
`:.y(0.2)=0.81902`
Method-2 : Using formula `k_2=hf(x_0+h/2,y_0+k_1/2)`
Second 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`
`y_1=y_0+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=hf(x_1,y_1)=(0.1)f(0.1,0.905)=(0.1)*(-0.905)=-0.0905`
`k_2=hf(x_1+h/2,y_1+k_1/2)=(0.1)f(0.15,0.85975)=(0.1)*(-0.85975)=-0.08598`
`y_2=y_1+k_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.
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