By Radhika, T.S.L

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Let us use examples to illustrate this method. 5), we obtain σ = 0. Thus, the solution for λ = i is ∞ y (x ) = e ix ∑c x n n=0 −n . Computing y ( x ) , y ( x ) and substituting in the given equation, the recurrence relation to find cn ’s is cn = n +1 cn−1 for n = 1, 2, 3 … 2i Thus, c1 = −ic 0 ; c 2 = 3 3 3 c1 = − c 0 ; c 3 = c1 = 3ic 0 2i 2 2i and so on. The solution is y ( x ) = e ix c 0 1 − ix −1 − 3 −2 x + 3ix −3 + 2 Similarly, for λ = −i, the solution is y ( x ) = e − ix c 0 1 + ix −1 − 3 −2 x − 3ix −3 + 2 53 A sy m p t o ti c Me t h o d Now, we can construct two real solutions to the given problem by defining u (x ) = y+ y 2 and v ( x ) = y− y .

Rudin. Principles of Mathematical Analysis. 2nd edition. New York: McGrawHill, 1976. F. Simmons. Differential Equations with Applications and Historical Notes. 2nd edition. Noida, India: Tata McGraw-Hill, 2003. M. Tenenbaum and H. Pollard. Ordinary Differential Equation. Mineola, NY: Dover, 1985. G. Zill. A First Course in Differential Equations with Modeling Applications. 9th edition. Stamford, CT: Brooks/Cole, Cengage Learning, 2008. 3 A symp toti c M e thod Introduction Chapter 2 discussed the power series method to find an approximate analytical solution to ordinary differential equations (ODEs).

19 To illustrate case (ii), consider the problem of finding two linearly independent solutions of x 2 y + xy + x 2 − 1 y=0 4 about x = 0. Solution: Here, x0 = 0 is an RSP of the given differential equation. 15) − an x n+m = 0 4 n=0 ∑ The indicial equation is m2 − 1 a0 = 0 4 which gives 1 m=− , 2 1 2 Observe that the two roots differ by an integer. Here, n = 1 and for m1 = 12 , the recurrence relation is an = − an− 2 , n = 2, 3, 4 n(n + 1) We see from this relation that a2 , a4 , a6 are in terms of a0 , whereas a3 , a5 , a7 ...