Find y(0.1) for `y''=-4z-4y`, `x_0=0, y_0=0, z_0=1`, with step length 0.1 using Taylor Series method (second order differential equation)

Solution:
Given `y^('')=-4z-4y, y(0)=0, y'(0)=1, h=0.1, y(0.1)=?`

put `(dy)/(dx)=z` and differentiate w.r.t. x, we obtain `(d^2y)/(dx^2)=(dz)/(dx)`

We have system of equations
`(dy)/(dx)=z=f(x,y,z)`

`(dz)/(dx)=-4z-4y=g(x,y,z)`

Here, `x_0=0,y_0=0,z_0=1,h=0.1,x_n=0.1`

Differentiating successively, we get

Derivative steps

`d/(dx)(-4y'-4y)`

`=-d/(dx)(4y')-d/(dx)(4y)`

`=-4y''-4y'`

Now, `d^2/(dx^2)(-4y'-4y)=d/(dx)(-4y''-4y')`

`=-d/(dx)(4y'')-d/(dx)(4y')`

`=-4y'''-4y''`

`y'=-4y'-4y`

`y''=-4y'-4y`

`y'''=-4y''-4y'`

`y^(iv)=-4y'''-4y''`

Now substituting, we get
`y_0''=-4y_0'-4y_0=-4`

`y_0'''=-4y_0''-4y_0'=12`

`y_0^(iv)=-4y_0'''-4y_0''=-32`

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=0,y_0'=1`

`=0+0.1*(1)+(0.1)^2/(2)*(-4)+(0.1)^3/(6)*(12)+(0.1)^4/(24)*(-32)+...`

`=0+0.1-0.02+0.002+0+...`

`=0.0819`

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

`:.y(0.1)=0.0819`

`n`	`x_n`	`y_n`	`y_n'`	`y_n''`	`y_n'''`	`y_n^(iv)`	`x_(n+1)`	`y_(n+1)`
0	0	0	1	-4	12	-32	0.1	0.0819

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