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differential equations

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x2 y''+xy'+(x2 - 1/4)y=0

i shifted the index by 2 and when i moved the other series up i end up with 2 sets of equations containing r and come up with 4 roots for some reason

Apr 2nd, 2015

Let a solution of the differential equation be represented as a series Σn=0 an xn+r , where a0 ≠ 0.

After substitution of the series into the differential equation we get

Σn=0 an [(r+n)2 – 1/4]xn+r + Σn=2 an–2 xn+r  = 0

If n = 0, then we have r2 – 1/4 = 0 and r = 1/2 or r = – 1/2.

If n = 1, then a1 = 0 because (r+1)2 – 1/4 ≠ 0. For n > 1 the coefficients can be found from the relations 

an = – an–2 /[(r+n)2 – 1/4] where r = 1/2 or r = – 1/2. Thus we get two linearly independent solutions of the differential equation. Its general solution can be obtained as their linear combination with arbitrary constants C1 and C2.


Apr 3rd, 2015

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