Formula

Taylor Series method `h=x-x_n` `y_(n+1) = y_n + hy_n' + h^2/(2!) y_n'' + h^3/(3!) y_n''' + h^4/(4!) y_n^(iv) + h^5/(5!) y_n^(v) + ...`

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Examples
1. Find y(0.1) for `y''=1+2xy-x^2z`, `x_0=0, y_0=1, z_0=0`, with step length 0.1 using Taylor Series method (second order differential equation)

Solution:
Given `y^('')=1+2xy-x^2z, y(0)=1, y'(0)=0, 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)=1+2xy-x^2z=g(x,y,z)`

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

Differentiating successively, we get

Derivative steps

`d/(dx)(1+2xy-x^2y')`

`=d/(dx)(1)+d/(dx)(2xy)-d/(dx)(x^2y')`

`d/(dx)(2xy)=2y+2xy'`

`d/(dx)(2xy)`

`=2*(d/(dx)(x))y+2x(d/(dx)(y))`

`=2*1y+2x(y')`

`=2y+2xy'`

`d/(dx)(x^2y')=2xy'+x^2y''`

`d/(dx)(x^2y')`

`=(d/(dx)(x^2))y'+x^2(d/(dx)(y'))`

`=(2x)y'+x^2(y'')`

`=2xy'+x^2y''`

`=0+(2y+2xy')-(2xy'+x^2y'')`

`=0+2y+2xy'-2xy'-x^2y''`

`=2y-x^2y''`

Now, `d^2/(dx^2)(1+2xy-x^2y')=d/(dx)(2y-x^2y'')`

`=d/(dx)(2y)-d/(dx)(x^2y'')`

`d/(dx)(x^2y'')=2xy''+x^2y'''`

`d/(dx)(x^2y'')`

`=(d/(dx)(x^2))y''+x^2(d/(dx)(y''))`

`=(2x)y''+x^2(y''')`

`=2xy''+x^2y'''`

`=2y'-(2xy''+x^2y''')`

`=2y'-2xy''-x^2y'''`

`y'=1+2xy-x^2y'`

`y''=1+2xy-x^2y'`

`y'''=2y-x^2y''`

`y^(iv)=2y'-2xy''-x^2y'''`

Now substituting, we get
`y_0''=1+2x_0y_0-x_0^2y_0'=1`

`y_0'''=2y_0-x_0^2y_0''=2`

`y_0^(iv)=2y_0'-2x_0y_0''-x_0^2y_0'''=0`

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,y_0'=0`

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

`=1+0+0.005+0+0+...`

`=1.0053`

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

`:.y(0.1)=1.0053`

`n`	`x_n`	`y_n`	`y_n'`	`y_n''`	`y_n'''`	`y_n^(iv)`	`x_(n+1)`	`y_(n+1)`
0	0	1	0	1	2	0	0.1	1.0053

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