For any integer k, write fk(x)=(1+x2)−1−k. When k≥1, find constants c1,c2 such that the function y=(Sfk)(x) solves a second order differential equation xy′′+c1y′+c2xy=0.
A number or a short expression. Spacing and $ signs are ignored.
Solution
Firstly, xjfk(x)(0≤j≤2k) are all absolutely integrable when k≥1. Then, by (6), y=(Sfk)(x) is a 2k-th order continuous differentiable function. By Lemma 0.1 and Lemma 0.4,xy′′+c1y′+c2xy=0 is equivalent to (x2fk′+2xfk)−c1xfk−4π2c2fk′=0. Inputting fk(x)=(1+x2)−1−k, we get c1=−2k and c2=−4π2.
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