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Algebra Difficulty 5.8 AIME, harder Find the answer

For any integer kk, write fk(x)=(1+x2)1kf_{k}(x)=\left(1+x^{2}\right)^{-1-k}. When k1k \geq 1, find constants c1,c2c_{1}, c_{2} such that the function y=(Sfk)(x)y=\left(S f_{k}\right)(x) solves a second order differential equation xy+c1y+c2xy=0x y^{\prime \prime}+c_{1} y^{\prime}+c_{2} x y=0.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Firstly, xjfk(x)(0j2k)x^{j} f_{k}(x)(0 \leq j \leq 2 k) are all absolutely integrable when k1k \geq 1. Then, by (6), y=(Sfk)(x)y=\left(S f_{k}\right)(x) is a 2k2 k-th order continuous differentiable function. By Lemma 0.1 and Lemma 0.4,xy+c1y+c2xy=00.4, x y^{\prime \prime}+c_{1} y^{\prime}+c_{2} x y=0 is equivalent to (x2fk+2xfk)c1xfkc24π2fk=0\left(x^{2} f_{k}^{\prime}+2 x f_{k}\right)-c_{1} x f_{k}-\frac{c_{2}}{4 \pi^{2}} f_{k}^{\prime}=0. Inputting fk(x)=(1+x2)1kf_{k}(x)=\left(1+x^{2}\right)^{-1-k}, we get c1=2kc_{1}=-2 k and c2=4π2c_{2}=-4 \pi^{2}.

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