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C nx n = 0 This simplifies to X ∞ n=0 (n(n − 1)4n2)c n −(n2)(n1)c n2xn = 0 14 From this we get the recurrence relation c n2 = (n2 3n2) (n2)(n1) c n = (n2)(n1) (n2)(n1) c n = c n So, we specify c0 and c1, and the rest of the coefficients are determined by the above recurrence relation This gives us a geometric series, and the corresponding solution to the ODE. J E Z O ̂ ƁA s N E ԁE E ΁E Ȃǂ̃e X g J g Ă Ȃ ̃x X g J O v f f A50 F قǂ̃p \ i J ̒ œ ɂ ̐F Đf f ܂ B. N @ _ g _ \ Z K l j Z g u P _ g l j Z e v g h c b < h k l h q g h c ?.

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J I f B I A J i r E j ^ ̎ t A ̔ B i C g C _ d l ̐ ܂ B i C g C _ d l ̎ԗ ̔ ǎ Ȓ ÎԂ 舵 Ă ܂ B C y ɂ k B z ̌` ƂȂ B g C N X uF6T v ́A x X ԗ Ƀz _ @ S h E B O ̗p B S Ƒ c ӎ ݌v A X ^ C b V ȃf U C ܂ B ւȂ ł͂̑u A ܂ l ւ̈ S ˔ A r W A ł͂Ȃ A u L Ȃǔ ɍS ƂȂ Ă ܂ B ؂ʼn K A ܂ ŐV ̋Z p g g C N \ B. T Z Z y5 N ۏ؁z G X H E H b ` U u b N/ V o f B X ~ j HH/UNO ̂ A ́A V ^ ̃K C h A ́A T Z Z y5 N ۏ؁z G X H E H b ` U u b N/ V o f B X ~ j HH/UNO Ɠ ̃R i ̒l ̕i A C e Q l ɂ́A ̍ i f C L O A ̏ i Љ T C g A ŐV G X f B X E H b ` ŏW ߂Ă 鏤 i ╨ B. C C } J n E X { b N X p e No3 iJAN R h j ̃y W ł B i 4 `5 c Ɠ ȓ ɔ ܂ i y j j B DCM I C ( ) n E X { b N X ̃p e w z Z ^ ʔ̃T C g ł BDCM I C ł͓h E C p i ͂ ߂Ƃ A 34 _ ̏ i 舵 Ă ܂ B z Z ^ ʔ DCM I C ł̂ y ݂ B.

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