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(514)-(517) we have f ′′ 1 (r) + 2 r f ′ 1(r)− 2f1(r) r2 = 0,(523) 2 r3 k1 = 0,(524) k2 +k 3 = 0.(525) However, sincek 1 ̸= 0, then Eq

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