2. Example-2
2. Find y(0.5) for `y'=-2x-y`, y(0) = -1, with step length 0.1 using Improved Euler method
Solution:
Given `y'=-2x-y, y(0)=-1, h=0.1, y(0.5)=?`
Here, `x_0=0,y_0=-1,h=0.1`
`y'=-2x-y`
`:. f(x,y)=-2x-y`
Improved Euler method
`y_(m+1)=y_m+1/2 h[f(x_m,y_m) + f(x_m+h,y_m + hf(x_m,y_m))]`
`f(x_0,y_0)=f(0,-1)=1`
`f(x_0+h,y_0 + hf(x_0,y_0))=f(0.1,-0.9)=0.7`
`y_1=y_0+1/2 h[f(x_0,y_0) + f(x_0+h,y_0 + hf(x_0,y_0))]`
`y_1=-1+0.1/2 * [1+0.7]=-0.915`
Again taking `(x_1,y_1)` in place of `(x_0,y_0)` repeat the process
`f(x_1,y_1)=f(0.1,-0.915)=0.715`
`f(x_1+h,y_1 + hf(x_1,y_1))=f(0.2,-0.8435)=0.4435`
`y_2=y_1+1/2 h[f(x_1,y_1) + f(x_1+h,y_1 + hf(x_1,y_1))]`
`y_2=-0.915+0.1/2 * [0.715+0.4435]=-0.85708`
Again taking `(x_2,y_2)` in place of `(x_1,y_1)` repeat the process
`f(x_2,y_2)=f(0.2,-0.85708)=0.45708`
`f(x_2+h,y_2 + hf(x_2,y_2))=f(0.3,-0.81137)=0.21137`
`y_3=y_2+1/2 h[f(x_2,y_2) + f(x_2+h,y_2 + hf(x_2,y_2))]`
`y_3=-0.85708+0.1/2 * [0.45708+0.21137]=-0.82365`
Again taking `(x_3,y_3)` in place of `(x_2,y_2)` repeat the process
`f(x_3,y_3)=f(0.3,-0.82365)=0.22365`
`f(x_3+h,y_3 + hf(x_3,y_3))=f(0.4,-0.80129)=0.00129`
`y_4=y_3+1/2 h[f(x_3,y_3) + f(x_3+h,y_3 + hf(x_3,y_3))]`
`y_4=-0.82365+0.1/2 * [0.22365+0.00129]=-0.81241`
Again taking `(x_4,y_4)` in place of `(x_3,y_3)` repeat the process
`f(x_4,y_4)=f(0.4,-0.81241)=0.01241`
`f(x_4+h,y_4 + hf(x_4,y_4))=f(0.5,-0.81117)=-0.18883`
`y_5=y_4+1/2 h[f(x_4,y_4) + f(x_4+h,y_4 + hf(x_4,y_4))]`
`y_5=-0.81241+0.1/2 * [0.01241-0.18883]=-0.82123`
`:.y(0.5)=-0.82123`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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