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Determin alower bound for the radius of convergence of series solutions about point x0: ((x^2)-2x-3)y'' +xy'+ 4y = 0; x0=4

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You need to divide the equation by coefficient of y", `x^2 -2x-3`  ,such that:

`y ''+ x/(x^2 -2x -3) *y' + 4y = 0`

You need to find the singular points, hence you should solve for x the equation`x^2 -2x -3 = 0`  such that:

`x^2 -2x -3 = 0`

`x^2 -2x - 2 - 1= 0 =gt (x^2-1) - 2(x+1) = 0`

`(x-1)(x+1+2) = 0 =gt x-1=0`

`x_1 = 1`

`x+3=0 =gt x_2=-3`

You need to determine the distance between `x_0=4`  and the singular point `x_1=1`  such that:

`d_1 = |1-4| =gt d=|-3| = 3`

`d_2 = |-3-4| =gt d = |-7| = 7`

Hence, the lower bound of radius of convergence is `min (d_1,d_2) = 3` .

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