In this section we showed that one solution of Bessel's equation of order
zero,
$$
L[y]=x^{2} y^{\prime \prime}+x y^{\prime}+x^{2} y=0
$$
is \(J_{0}\), where \(J_{0}(x)\) is given by Fa. ( 7) with \(a_{0}=1\). According to
Theorem 5.7 .1 a second solution has the form \((x>0)\)
$$
y_{2}(x)=J_{0}(x) \ln x+\sum_{n=1}^{\infty} b_{n} x^{n}
$$
(a) Show that
$$
L\left[y_{2}\right](x)=\sum_{n=2}^{\infty} n(n-1) b_{n}
x^{n}+\sum_{n=1}^{\infty} n b_{n} x^{n}+\sum_{n=1}^{\infty} b_{n} x^{n+2}+2 x
J_{0}^{\prime}(x)
$$
(b) Substituting the series representation for \(J_{0}(x)\) in Eq. (i), show
that
$$
b_{1} x+2^{2} b_{2} x^{2}+\sum_{n=3}^{\infty}\left(n^{2} b_{n}+b_{n-2}\right)
x^{n}=-2 \sum_{n=1}^{\infty} \frac{(-1)^{n} 2 n x^{2 n}}{2^{2 n}(n !)^{2}}
$$
(c) Note that only even powers of \(x\) appear on the right side of Eq. (ii).
Show that \(b_{1}=b_{3}=b_{5}=\cdots=0, b_{2}=1 / 2^{2}(1 !)^{2},\) and that
$$
(2 n)^{2} b_{2 n}+b_{2 n-2}=-2(-1)^{n}(2 n) / 2^{2 n}(n !)^{2}, \quad n=2,3,4,
\ldots
$$
Deduce that
$$
b_{4}=-\frac{1}{2^{2} 4^{2}}\left(1+\frac{1}{2}\right) \quad \text { and }
\quad b_{6}=\frac{1}{2^{2} 4^{2} 6^{2}}\left(1+\frac{1}{2}+\frac{1}{3}\right)
$$
The general solution of the recurrence relation is \(b_{2 n}=(-1)^{n+1} H_{n} /
2^{2 n}(n !)^{2}\). Substituting for \(b_{n}\) in the expression for \(y_{2}(x)\)
we obtain the solution given in \(\mathrm{Eq} .(10) .\)
