Find y(0.2) for `y'=-y`, `x_0=0, y_0=1`, with step length 0.1 using Taylor Series method (first order differential equation)

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^(iv)=-y'''`

Now substituting, we get
`y_0'=-y_0=-1`

`y_0''=-y_0'=1`

`y_0'''=-y_0''=-1`

`y_0^(iv)=-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^(iv) + ...`

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for `n=0,x_0=0,y_0=1`

`=1+0.1*(-1)+(0.1)^2/(2)*(1)+(0.1)^3/(6)*(-1)+(0.1)^4/(24)*(1)+...`

`=1-0.1+0.005+0+0+...`

`=0.9048`

`x_1=x_0+h=0+0.1=0.1`

Now substituting, we get
`y_1'=-y_1=-0.9048`

`y_1''=-y_1'=0.9048`

`y_1'''=-y_1''=-0.9048`

`y_1^(iv)=-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^(iv) + ...`

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for `n=1,x_1=0.1,y_1=0.9048`

`=0.9048+0.1*(-0.9048)+(0.1)^2/(2)*(0.9048)+(0.1)^3/(6)*(-0.9048)+(0.1)^4/(24)*(0.9048)+...`

`=0.9048-0.0905+0.0045+0+0+...`

`=0.8187`

`x_2=x_1+h=0.1+0.1=0.2`

`:.y(0.2)=0.8187`

`n`	`x_n`	`y_n`	`y_n'`	`y_n''`	`y_n'''`	`y_n^(iv)`	`x_(n+1)`	`y_(n+1)`
0	0	1	-1	1	-1	1	0.1	0.9048
1	0.1	0.9048	-0.9048	0.9048	-0.9048	0.9048	0.2	0.8187

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