5. Example-2
Find y(0.5) for `y'=-2x-y`, `x_0=0, y_0=-1`, with step length 0.1 using Runge-Kutta 2 method (1st order derivative)
Solution:
Given `y'=-2x-y, y(0)=-1, h=0.1, y(0.5)=?`
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)=0.7`
`y_1=y_0+h/2(k_1+k_2)=-1+0.085=-0.915`
`:.y(0.1)=-0.915`
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,-0.915)=0.715`
`k_2=f(x_1+h,y_1+hk_1)=f(0.2,-0.8435)=0.4435`
`y_2=y_1+h/2(k_1+k_2)=-0.915+0.05792=-0.85708`
`:.y(0.2)=-0.85708`
Again taking `(x_2,y_2)` in place of `(x_1,y_1)` and repeat the process
`k_1=f(x_2,y_2)=f(0.2,-0.85708)=0.45708`
`k_2=f(x_2+h,y_2+hk_1)=f(0.3,-0.81137)=0.21137`
`y_3=y_2+h/2(k_1+k_2)=-0.85708+0.03342=-0.82365`
`:.y(0.3)=-0.82365`
Again taking `(x_3,y_3)` in place of `(x_2,y_2)` and repeat the process
`k_1=f(x_3,y_3)=f(0.3,-0.82365)=0.22365`
`k_2=f(x_3+h,y_3+hk_1)=f(0.4,-0.80129)=0.00129`
`y_4=y_3+h/2(k_1+k_2)=-0.82365+0.01125=-0.81241`
`:.y(0.4)=-0.81241`
Again taking `(x_4,y_4)` in place of `(x_3,y_3)` and repeat the process
`k_1=f(x_4,y_4)=f(0.4,-0.81241)=0.01241`
`k_2=f(x_4+h,y_4+hk_1)=f(0.5,-0.81117)=-0.18883`
`y_5=y_4+h/2(k_1+k_2)=-0.81241-0.00882=-0.82123`
`:.y(0.5)=-0.82123`
`:.y(0.5)=-0.82123`
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)=0.85`
`y_1=y_0+hk_2=-1+0.085=-0.915`
`:.y(0.1)=-0.915`
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,-0.915)=0.715`
`k_2=f(x_1+h/2,y_1+(hk_1)/2)=f(0.15,-0.87925)=0.57925`
`y_2=y_1+hk_2=-0.915+0.05792=-0.85708`
`:.y(0.2)=-0.85708`
Again taking `(x_2,y_2)` in place of `(x_1,y_1)` and repeat the process
`k_1=f(x_2,y_2)=f(0.2,-0.85708)=0.45708`
`k_2=f(x_2+h/2,y_2+(hk_1)/2)=f(0.25,-0.83422)=0.33422`
`y_3=y_2+hk_2=-0.85708+0.03342=-0.82365`
`:.y(0.3)=-0.82365`
Again taking `(x_3,y_3)` in place of `(x_2,y_2)` and repeat the process
`k_1=f(x_3,y_3)=f(0.3,-0.82365)=0.22365`
`k_2=f(x_3+h/2,y_3+(hk_1)/2)=f(0.35,-0.81247)=0.11247`
`y_4=y_3+hk_2=-0.82365+0.01125=-0.81241`
`:.y(0.4)=-0.81241`
Again taking `(x_4,y_4)` in place of `(x_3,y_3)` and repeat the process
`k_1=f(x_4,y_4)=f(0.4,-0.81241)=0.01241`
`k_2=f(x_4+h/2,y_4+(hk_1)/2)=f(0.45,-0.81179)=-0.08821`
`y_5=y_4+hk_2=-0.81241-0.00882=-0.82123`
`:.y(0.5)=-0.82123`
`:.y(0.5)=-0.82123`
This material is intended as a summary. Use your textbook for detail explanation.
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