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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Luca B.
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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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