2. Example-2
Find y(0.5) for `y'=-2x-y`, `x_0=0, y_0=-1`, with step length 0.1 using Runge-Kutta 4 method (1st order derivative)
Solution:
Given `y'=-2x-y, y(0)=-1, h=0.1, y(0.5)=?`
Forth 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)*(0.85)=0.085`
`k_3=hf(x_0+h/2,y_0+k_2/2)=(0.1)f(0.05,-0.9575)=(0.1)*(0.8575)=0.08575`
`k_4=hf(x_0+h,y_0+k_3)=(0.1)f(0.1,-0.91425)=(0.1)*(0.71425)=0.07142`
`y_1=y_0+1/6(k_1+2k_2+2k_3+k_4)`
`y_1=-1+1/6[0.1+2(0.085)+2(0.08575)+(0.07142)]`
`y_1=-0.91451`
`:.y(0.1)=-0.91451`
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,-0.91451)=(0.1)*(0.71451)=0.07145`
`k_2=hf(x_1+h/2,y_1+k_1/2)=(0.1)f(0.15,-0.87879)=(0.1)*(0.57879)=0.05788`
`k_3=hf(x_1+h/2,y_1+k_2/2)=(0.1)f(0.15,-0.88557)=(0.1)*(0.58557)=0.05856`
`k_4=hf(x_1+h,y_1+k_3)=(0.1)f(0.2,-0.85596)=(0.1)*(0.45596)=0.0456`
`y_2=y_1+1/6(k_1+2k_2+2k_3+k_4)`
`y_2=-0.91451+1/6[0.07145+2(0.05788)+2(0.05856)+(0.0456)]`
`y_2=-0.85619`
`:.y(0.2)=-0.85619`
Again taking `(x_2,y_2)` in place of `(x_0,y_0)` and repeat the process
`k_1=hf(x_2,y_2)=(0.1)f(0.2,-0.85619)=(0.1)*(0.45619)=0.04562`
`k_2=hf(x_2+h/2,y_2+k_1/2)=(0.1)f(0.25,-0.83338)=(0.1)*(0.33338)=0.03334`
`k_3=hf(x_2+h/2,y_2+k_2/2)=(0.1)f(0.25,-0.83952)=(0.1)*(0.33952)=0.03395`
`k_4=hf(x_2+h,y_2+k_3)=(0.1)f(0.3,-0.82224)=(0.1)*(0.22224)=0.02222`
`y_3=y_2+1/6(k_1+2k_2+2k_3+k_4)`
`y_3=-0.85619+1/6[0.04562+2(0.03334)+2(0.03395)+(0.02222)]`
`y_3=-0.82246`
`:.y(0.3)=-0.82246`
Again taking `(x_3,y_3)` in place of `(x_0,y_0)` and repeat the process
`k_1=hf(x_3,y_3)=(0.1)f(0.3,-0.82246)=(0.1)*(0.22246)=0.02225`
`k_2=hf(x_3+h/2,y_3+k_1/2)=(0.1)f(0.35,-0.81133)=(0.1)*(0.11133)=0.01113`
`k_3=hf(x_3+h/2,y_3+k_2/2)=(0.1)f(0.35,-0.81689)=(0.1)*(0.11689)=0.01169`
`k_4=hf(x_3+h,y_3+k_3)=(0.1)f(0.4,-0.81077)=(0.1)*(0.01077)=0.00108`
`y_4=y_3+1/6(k_1+2k_2+2k_3+k_4)`
`y_4=-0.82246+1/6[0.02225+2(0.01113)+2(0.01169)+(0.00108)]`
`y_4=-0.81096`
`:.y(0.4)=-0.81096`
Again taking `(x_4,y_4)` in place of `(x_0,y_0)` and repeat the process
`k_1=hf(x_4,y_4)=(0.1)f(0.4,-0.81096)=(0.1)*(0.01096)=0.0011`
`k_2=hf(x_4+h/2,y_4+k_1/2)=(0.1)f(0.45,-0.81041)=(0.1)*(-0.08959)=-0.00896`
`k_3=hf(x_4+h/2,y_4+k_2/2)=(0.1)f(0.45,-0.81544)=(0.1)*(-0.08456)=-0.00846`
`k_4=hf(x_4+h,y_4+k_3)=(0.1)f(0.5,-0.81942)=(0.1)*(-0.18058)=-0.01806`
`y_5=y_4+1/6(k_1+2k_2+2k_3+k_4)`
`y_5=-0.81096+1/6[0.0011+2(-0.00896)+2(-0.00846)+(-0.01806)]`
`y_5=-0.81959`
`:.y(0.5)=-0.81959`
`:.y(0.5)=-0.81959`
This material is intended as a summary. Use your textbook for detail explanation.
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