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By S S Cheng

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**Sample text**

12) which has a regular singular point at t = 0. 7. 1) into the form 2 d2 y 1 1 1 dy 1 − 2 p1 + 4 p2 + y = 0. 13). Use this substitution to show that for the DE. y ′′ + 1 2 1 1 + 2 x x y′ + 1 y=0 2x3 the point x = ∞ is a regular singular point. 8. 15) the point x = ∞ is an irregular singular point. 9. Examine the nature of the point at infinity for the following DEs: Airy’s DE: y ′′ − xy = 0 Chebyshev’s DE: (1 − x2 )y ′′ − xy ′ + a2 y = 0 Hermite’s DE: y ′′ − 2xy ′ + 2ay = 0 Hypergeometric DE: x(1 − x)y ′′ + [c − (a + b + 1)x]y ′ − aby = 0 Laguerre’s DE: xy ′′ + (a + 1 − x)y ′ + by = 0 Legendre’s DE: (1 − x2 )y ′′ − 2xy ′ + a(a + 1)y = 0.

DEs: (i) Compute the indicial equation and their roots for the following 2xy ′′ + y ′ + xy = 0 42 Lecture 6 (ii) x2 y ′′ + xy ′ + (x2 − 1/9)y = 0 (iii) x2 y ′′ + (x + x2 )y ′ − y = 0 (iv) x2 y ′′ + xy ′ + (x2 − 1/4)y = 0 (v) x(x − 1)y ′′ + (2x − 1)y ′ − 2y = 0 (vi) x2 y ′′ + 3 sin xy ′ − 2y = 0 (vii) x2 y ′′ + (1/2)(x + sin x)y ′ + y = 0 (viii) x2 y ′′ + xy ′ + (1 − x)y = 0. 2. Verify that each of the given DEs has a regular singular point at the indicated point x = x0 , and express their solutions in terms of power series valid for x > x0 : (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) (xii) (xiii) (xiv) (xv) 4xy ′′ + 2y ′ + y = 0, x = 0 9x2 y ′′ + 9xy ′ + (9x2 − 1)y = 0, x = 0 2x2 y ′′ + xy ′ − (x + 1)y = 0, x = 0 (1 − x2 )y ′′ + y ′ + 2y = 0, x = −1 x2 y ′′ + (x2 − 7/36)y = 0, x = 0 x2 y ′′ + (x2 − x)y ′ + 2y = 0, x = 0 x2 y ′′ + (x2 − x)y ′ + y = 0, x = 0 x(1 − x)y ′′ + (1 − 5x)y ′ − 4y = 0, x = 0 (x2 + x3 )y ′′ − (x + x2 )y ′ + y = 0, x = 0 x2 y ′′ + 2xy ′ + xy = 0, x = 0 x2 y ′′ + 4xy ′ + (2 + x)y = 0, x = 0 x(1 − x)y ′′ − 3xy ′ − y = 0, x = 0 x2 y ′′ − (x + 2)y = 0, x = 0 x(1 + x)y ′′ + (x + 5)y ′ − 4y = 0, x = 0 (x − x2 )y ′′ − 3y ′ + 2y = 0, x = 0.

6) reduces to (m + r)(m + r + 1)cm = cm−1 , m = 1, 2, · · · which easily gives cm = 1 c0 , (r + 1)(r + 2)2 (r + 3)2 · · · (r + m)2 (r + m + 1) m = 1, 2, · · · . 7) reduces to cm = 1 c0 , m! (m + 1)! 7) m = 1, 2, · · · ; therefore, the first solution y1 (x) is given by y1 (x) = ∞ 1 xm . m! (m + 1)! 7) is the same as cm = 1 , (r + 2)2 · · · (r + m)2 (r + m + 1) m = 1, 2, · · · . ) 41 and hence e0 = c′0 (−1) = em = c′m (−1) = = 1 1 −2 2 2 1 · 2 · · · (m − 1)2 m − 1 2 m! (m − 1)! m−1 k=1 m−1 k=1 1 1 − k m 1 1 + , k m m = 1, 2, · · · .