Find y(0.2) for `y'=-y`, `x_0=0, y_0=1`, with step length 0.1 using Taylor Series 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`
Differentiating successively, we get
Derivative steps
`d/(dx)(-y)`
`=-y'`
Now, `d^2/(dx^2)(-y)=d/(dx)(-y')`
`=-y''`
Now, `d^3/(dx^3)(-y)=d/(dx)(-y'')`
`=-y'''`
`y'=-y`
`y''=-y'`
`y'''=-y''`
`y''''=-y'''`
Now substituting, we get
`y_0'=-y_0=-1`
`y_0''=-y_0'=1`
`y_0'''=-y_0''=-1`
`y_0''''=-y_0'''=1`
Putting these values in Taylor Series, we have
`y_1 = y_0 + hy_0' + h^2/(2!) y_0'' + h^3/(3!) y_0''' + h^4/(4!) y_0'''' + ...`
`=1+0.1*(-1)+(0.1)^2/(2!)*(1)+(0.1)^3/(3!)*(-1)+(0.1)^4/(4!)*(1)+...`
`=1-0.1+0.005+0+0+...`
`=0.9048`
`:.y(0.1)=0.9048`
Again taking `(x_1,y_1)` in place of `(x_0,y_0)` and repeat the process
Now substituting, we get
`y_1'=-y_1=-0.9048`
`y_1''=-y_1'=0.9048`
`y_1'''=-y_1''=-0.9048`
`y_1''''=-y_1'''=0.9048`
Putting these values in Taylor Series, we have
`y_2 = y_1 + hy_1' + h^2/(2!) y_1'' + h^3/(3!) y_1''' + h^4/(4!) y_1'''' + ...`
`=0.9048+0.1*(-0.9048)+(0.1)^2/(2!)*(0.9048)+(0.1)^3/(3!)*(-0.9048)+(0.1)^4/(4!)*(0.9048)+...`
`=0.9048-0.0905+0.0045+0+0+...`
`=0.8187`
`:.y(0.2)=0.8187`
This material is intended as a summary. Use your textbook for detail explanation.
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