3. Example-3
Find y(0.2) for `y'=-y`, `x_0=0, y_0=1`, with step length 0.1 using Modified Euler method (1st order derivative)
Solution:
Given `y'=-y, y(0)=1, h=0.1, y(0.2)=?`
Here, `x_0=0,y_0=1,h=0.1,x_n=0.2`
`y'=-y`
`:. f(x,y)=-y`
Modified Euler method
`y_(m+1)=y_m+hf(x_m+1/2 h,y_m + 1/2 hf(x_m,y_m))`
`f(x_0,y_0)=f(0,1)=-1`
`x_0+1/2 h=0+0.1/2 =0.05`
`y_0 + 1/2 hf(x_0,y_0)=1+0.1/2 * -1=0.95`
`f(x_0+1/2 h, y_0 + 1/2 hf(x_0,y_0)=f(0.05,0.95)=-0.95`
`y_1=y_0+hf(x_0+1/2 h, y_0 + 1/2 hf(x_0,y_0))=1+0.1*-0.95=0.905`
`:.y(0.1)=0.905`
Again taking `(x_1,y_1)` in place of `(x_0,y_0)` and repeat the process
`f(x_1,y_1)=f(0.1,0.905)=-0.905`
`x_1+1/2 h=0.1+0.1/2 =0.15`
`y_1 + 1/2 hf(x_1,y_1)=0.905+0.1/2 * -0.905=0.8598`
`f(x_1+1/2 h, y_1 + 1/2 hf(x_1,y_1)=f(0.15,0.8598)=-0.8598`
`y_2=y_1+hf(x_1+1/2 h, y_1 + 1/2 hf(x_1,y_1))=0.905+0.1*-0.8598=0.819`
`:.y(0.2)=0.819`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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