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We showed that it can always be written as a linear combination of the two independent solutions y1 and y2 , at least within the range of x for which they have convergent power-series expansions. Therefore there are exactly two linearly independent solutions. It is clear that very similar arguments could be used for an N ’th-order ODE, and would show that it has N linearly-independent solutions. 11), or L(y) = 0, for which each term is of degree 1 in y or its derivatives. e. L(y) ≡ y ′′ (x) + p(x) y ′ (x) + q(x) y(x) = r(x) .

75). Let us suppose that x = a is a regular singular point. This corresponds to the case where p(x) has a pole of order 1, and q(x) has a pole of order 2. From the previous discussion, we are expecting that the solution will have singular behaviour of the general form (x − a)s . 77) for some constant index s. 76) is given by J = q(x) + s p(x) s(s − 1) + . 79) where F (x) and G(x) are analytic. 80) and so J= G(a) + s F (a) + s (s − 1) G′ (a) + s F ′ (a) + + regular terms . 81) Assume for a moment that G′ (a) + s F ′ (a) = 0, so that there is no 1/(x − a) term.

When s1 − s2 is not equal to an integer, this will give y2 (x) = u1 (x) bn (x − a)n+s2 . 83). 91) may not be literally identical; the one may be related to the other by a constant scaling and the addition of some constant multiple of y1 (x). 83) is guaranteed to be linearly independent of y1 (x). 91) is also guaranteed to be linearly independent of y1 (x). It is to be hoped that no undue confusion has been casued by giving the results of these two constructions for the second solution the same name y2 (x).