I am reading a textbook about engineering mathematics. To solve the Bessel differential equation $$ x^2 y'' + xy' + (x^2 - \nu^2) y = 0 $$ the book first obtains the function $J_{n}(x)$ using the series method as follows: $$ J_n(x) = x^n \sum_{m=0}^{\infty} \frac{(-1)^m x^{2m}}{2^{2m+n} m! (n+m)!} $$
then obtains the Bessel function of the second kind of zero order $Y_{0}(x)$ using the series method as follows: $$ Y_{0}(x) = \frac{2}{\pi} \left[ J_0(x) \left( \ln\frac{x}{2} + \gamma \right) + \sum_{m=1}^{\infty} \frac{(-1)^{m-1} h_m}{2^{2m}(m!)^2} x^{2m} \right] $$ in which $$ h_m = 1 + \frac12 + \cdots + \frac1m $$ and $\gamma \simeq 0.5772$ is the so-called ''Euler constant'', which is defined as the limit of $$ 1 + \frac12 + \cdots + \frac1s - \ln s $$ as $s$ approaches infinity. Then the author states that the Bessel function of the second kind of order $n$ is obtained in a similar way as: \begin{align} Y_{n}(x) &= \frac{2}{\pi} J_{n}(x) \left(\ln\frac{x}{2} + \gamma \right) \\[2pt] &\quad{} + \frac{x^n}{\pi} \sum_{m=0}^{\infty} \frac{(-1)^{m-1}(h_{m} + h_{m+n})}{2^{2m+n} m! (m+n)!} x^{2m} \\ &\quad{} - \frac{x^{-n}}{\pi} \sum_{m=0}^{n-1} \frac{(n-m-1)!}{2^{2m-n} m!} x^{2m} \end{align} The book does not provide any proof for the latter formula. Can you explain me how to get it? The mentioned book is "ADVANCED ENGINEERING MATHEMATICS" by "ERWIN KREYSZIG". If you know a textbook that explains how to get this formula well, please introduce it to me
Edit: I found a solution myself, but I'm not sure if the details are correct. here is the solution: According to the Frobenius method, a second solution is obtained from this formula: $${\it y}_{{\rm 2}}{\rm (}{\it x}{\rm )}={\it kJ}_{{\it n}}{\rm (}{\it x}{\rm )}{\it Lnx}+{\it x}^{{\rm -}{\it n}}\sum\limits_{{\it m}={\rm 0}}^{\infty}{{\it a}_{{\it m}}{\it x}^{{\it m}}}$$ By differentiating this function twice, we get $y{_{2}}^{\prime}$ and $y{_{2}}^{\prime\prime}$ $$y{_{2}}^{\prime}(x)=kJ{_{n}}^{\prime}(x)Lnx+{{KJ_{n}(x)}\over{x}}+\sum\limits_{m=0}^{\infty}{(m-n)a_{m}x^{m-n-1}}$$ $$y{_{2}}^{\prime\prime}(x)=kJ{_{n}}^{\prime\prime}(x)Lnx+{{2KJ{_{n}}^{\prime}(x)}\over{x}}-{{KJ_{n}(x)}\over{x^{2}}}+\sum\limits_{m=0}^{\infty}{(m-n)(m-n-1)a_{m}x^{m-n-2}}$$ Substituting ${\it y}_{{\rm 2}}{\rm (}{\it x}{\rm )}$, $y{_{2}}^{\prime}(x)$, and $y{_{2}}^{\prime\prime}(x)$ into Bessel equation $x^{2}y''+xy'+(x^{2}-\upsilon^{2})y=0$, we get $$kx^{2}J{_{n}}^{\prime}(x)Ln(x)+2kxJ{_{n}}^{\prime}(x)+kxJ{_{n}}^{\prime}(x)Ln(x)+k(x^{2}-n^{2})J_{n}(x)Ln(x)$$ $$+\sum\limits_{{\it m}={\rm 0}}^{\infty}{{\rm (}{\it m}-{\it n}{\rm )(}{\it m}-{\it n}-{\rm 1}{\rm )}{\it a}_{{\it m}}{\it x}^{{\it m}{\rm -}{\it n}}}+\sum\limits_{{\it m}{\rm =0}}^{{\rm \infty}}{{\rm (}{\it m}-{\it n}{\rm )}}{\it a}_{{\it m}}{\it x}^{{\it m}{\rm -}{\it n}}+\sum\limits_{{\it m}={\rm 0}}^{\infty}{{\it a}_{{\it m}}}{\it x}^{{\it m}{\rm -}{\it n}{\rm +2}}-{\it n}^{{\rm 2}}\sum\limits_{{\it m}={\rm 0}}^{\infty}{{\it a}_{{\it m}}}{\it x}^{{\it m}{\rm -}{\it n}}={\rm 0}$$ $$\Rightarrow 2kxJ{_{n}}^{\prime}(x)+\sum\limits_{m=0}^{\infty}{\left[{(m-n)^{2}-n^{2}}\right]}a_{m}x^{m-n}+\sum\limits_{m=0}^{\infty}{a_{m}}x^{m-n+2}=0$$ or $$\Rightarrow 2kxJ{_{n}}^{\prime}(x)+\sum\limits_{m=0}^{\infty}{m(m-2n)}a_{m}x^{m-n}+\sum\limits_{m=0}^{\infty}{a_{m}}x^{m-n+2}=0\ \ \ \ \ \ (1)$$ find the derivative of ${\it J}_{{\it n}}{\rm (}{\it x}{\rm )}$ $$J_{n}(x)=\sum\limits_{m=0}^{\infty}{{{(-1)^{m}}\over{2^{2m+n}m!(m+n)!}}}x^{2m+n}\Rightarrow J{_{n}}^{\prime}(x)=\sum\limits_{m=0}^{\infty}{{{(-1)^{m}(2m+n)}\over{2^{2m+n}m!(m+n)!}}}x^{2m+n-1}$$ Substituting in $(1)$ we have: $${\rm 2}{\it k}\sum\limits_{{\it m}{\rm =0}}^{{\rm \infty}}{{{{\rm (}-{\rm 1}{\rm )}^{{\it m}}{\rm (}{\rm 2}{\it m}+{\it n}{\rm )}}\over{{\rm 2}^{{\rm 2}{\it m}{\rm +}{\it n}}{\it m}{\rm !(}{\it m}+{\it n}{\rm )!}}}}{\it x}^{{\rm 2}{\it m}{\rm +}{\it n}}+\sum\limits_{{\it m}={\rm 0}}^{\infty}{{\it m}{\rm (}{\it m}-{\rm 2}{\it n}{\rm )}{\rm a}_{{\it m}}{\it x}^{{\it m}{\rm -}{\it n}}}{\rm }+\sum\limits_{{\it m}={\rm 0}}^{\infty}{{\it a}_{{\it m}}}{\it x}^{{\it m}{\rm -}{\it n}{\rm +2}}={\rm 0}$$ For the coefficient of $x^{-n}$: $0a_{0}=0$ so $a_{0}$ is an arbitrary constant
For the coefficient of $x^{1-n}$: $1(1-2n)a_{1}=0\Rightarrow a_{1}=0$
And in general for the coefficient of $x^{p-n+2}$ $(p\leq 2n-3)$:
$${\rm (}{\it p}+{\rm 2}{\rm )(}{\it p}+{\rm 2}-{\rm 2}{\it n}{\rm )}{\it a}_{{\it p}{\rm +2}}+{\it a}_{{\it p}}={\rm 0}\Rightarrow{\it a}_{{\it p}{\rm +2}}=-{{{\it a}_{{\it p}}}\over{{\rm (}{\it p}+{\rm 2}{\rm )(}{\it p}+{\rm 2}-{\rm 2}{\it n}{\rm )}}}$$
And according to this recursive relation, we have:
$${\it a}_{{\rm 3}}={\it a}_{{\rm 5}}=\cdots={\it a}_{{\rm 2}{\it n}{\rm -1}}={\rm 0}$$
That is, all the odd coefficients up to ${\it a}_{{\rm 2}{\it n}{\rm -1}}$ are equal to zero, but this result is also true for the odd coefficients greater than ${\it a}_{{\rm 2}{\it n}{\rm -1}}$, because if the power of $x$ is $2k+n$, ${\it m}$ is $2n+2k$ in the second sigma and $2n+2k-2$ in the third sigma, that is, the even coefficients of ${\it a}$ are produced So the equation
$${\it a}_{{\it p}{\rm +2}}=-{{{\it a}_{{\it p}}}\over{{\rm (}{\it p}+{\rm 2}{\rm )(}{\it p}+{\rm 2}-{\rm 2}{\it n}{\rm )}}}$$
holds for all odd coefficients and we can conclude that all odd coefficients are equal to zero and we only calculate even coefficients.
$${\it a}_{{\rm 2}}=-{{{\it a}_{{\rm 0}}}\over{{\rm 2}{\rm (}{\rm 2}-{\rm 2}{\it n}{\rm )}}}={{{\it a}_{{\rm 0}}}\over{{\rm 2}{\rm (}{\rm 2}{\it n}-{\rm 2}{\rm )}}}$$
$${\it a}_{{\rm 4}}=-{{{\it a}_{{\rm 2}}}\over{{\rm 4}{\rm (}{\rm 4}-{\rm 2}{\it n}{\rm )}}}={{{\it a}_{{\rm 2}}}\over{{\rm 4}{\rm (}{\rm 2}{\it n}-{\rm 4}{\rm )}}}={{{\it a}_{{\rm 0}}}\over{{\rm 4}\times{\rm 2}{\rm (}{\rm 2}{\it n}-{\rm 2}{\rm )(}{\rm 2}{\it n}-{\rm 4}{\rm )}}}$$
$${\it a}_{{\rm 6}}=-{{{\it a}_{{\rm 4}}}\over{{\rm 6}{\rm (}{\rm 6}-{\rm 2}{\it n}{\rm )}}}={{{\it a}_{{\rm 4}}}\over{{\rm 6}{\rm (}{\rm 2}{\it n}-{\rm 6}{\rm )}}}={{{\it a}_{{\rm 0}}}\over{{\rm 6}\times{\rm 4}\times{\rm 2}{\rm (}{\rm 2}{\it n}-{\rm 2}{\rm )(}{\rm 2}{\it n}-{\rm 4}{\rm )(}{\rm 2}{\it n}-{\rm 6}{\rm )}}}$$
and in general for $k\leq n-1$:
$${\it a}_{{\rm 2}{\it k}}={{{\it a}_{{\rm 0}}}\over{{\rm 2}{\it k}{\rm (}{\rm 2}{\it k}-{\rm 2}{\rm )}\cdots\times{\rm 4}\times{\rm 2}{\rm (}{\rm 2}{\it n}-{\rm 2}{\rm )(}{\rm 2}{\it n}-{\rm 4}{\rm )}\cdots{\rm (}{\rm 2}{\it n}-{\rm 2}{\it k}{\rm )}}}={{{\it a}_{{\rm 0}}}\over{{\rm 2}^{{\it k}}\times{\it k}{\rm !}\times{\rm 2}^{{\it k}}{\rm (}{\it n}-{\rm 1}{\rm )}\cdots{\rm (}{\it n}-{\it k}{\rm )}}}$$
$$={{{\it a}_{{\rm 0}}}\over{{\rm 2}^{{\rm 2}{\it k}}\times{\it k}{\rm !(}{\it n}-{\rm 1}{\rm )}\cdots{\rm (}{\it n}-{\it k}{\rm )}}}={{{\it a}_{{\rm 0}}}\over{{\rm 2}^{{\rm 2}{\it k}}\times{\it k}{\rm !}{{{\rm (}{\it n}-{\rm 1}{\rm )!}}\over{{\rm (}{\it n}-{\it k}-{\rm 1}{\rm )!}}}}}={{{\rm (}{\it n}-{\it k}-{\rm 1}{\rm )!}{\it a}_{{\rm 0}}}\over{{\rm 2}^{{\rm 2}{\it k}}\times{\it k}{\rm !(}{\it n}-{\rm 1}{\rm )!}}}$$
We assume that ${\it a}_{{\rm 0}}=-{\rm 2}^{{\it n}{\rm -1}}{\rm (}{\it n}-{\rm 1}{\rm )!}$. with this assumption, the coefficients will be in this form:
$${\it a}_{{\rm 2}{\it k}}={{-{\rm 2}^{{\it n}{\rm -1}}{\rm (}{\it n}-{\rm 1}{\rm )!(}{\it n}-{\it k}-{\rm 1}{\rm )!}}\over{{\rm 2}^{{\rm 2}{\it k}}\times{\it k}{\rm !(}{\it n}-{\rm 1}{\rm )!}}}=-{{{\rm (}{\it n}-{\it k}-{\rm 1}{\rm )!}}\over{{\rm 2}^{{\rm 2}{\it k}{\rm -}{\it n}{\rm +1}}{\it k}{\rm !}}}{\rm (}I{\rm )}$$
In particular ${\it a}_{{\rm 2}{\it n}{\rm -2}}=-{{{\rm 1}}\over{{\rm 2}^{{\it n}{\rm -1}}{\rm (}{\it n}-{\rm 1}{\rm )!}}}$.
For the coefficient of $x^{n}$: $${\rm 2}{\it k}{{{\it n}}\over{{\rm 2}^{{\it n}}{\it n}{\rm !}}}+{\rm 0}{\it a}_{{\rm 2}{\it n}}+{\it a}_{{\rm 2}{\it n}{\rm -2}}={\rm 0}\Rightarrow{{{\it k}}\over{{\rm 2}^{{\it n}{\rm -1}}{\rm (}{\it n}-{\rm 1}{\rm )!}}}-{{{\rm 1}}\over{{\rm 2}^{{\it n}{\rm -1}}{\rm (}{\it n}-{\rm 1}{\rm )!}}}={\rm 0}$$ $$\Rightarrow k=1$$ The above relationship holds for all values of ${\it a}_{{\rm 2}{\it n}}$, so ${\it a}_{{\rm 2}{\it n}}$ is an arbitrary parameter and we can assume that $${\it a}_{{\rm 2}{\it n}}=-{{{\rm 1}+{{{\rm 1}}\over{{\rm 2}}}+\cdots+{{{\rm 1}}\over{{\it n}}}}\over{{\rm 2}^{{\it n}{\rm +1}}{\it n}{\rm !}}}$$ For the coefficient of $x^{2+n}$: $${\rm 2}{{{\rm (}-{\rm 1}{\rm )(}{\rm 2}+{\it n}{\rm )}}\over{{\rm 2}^{{\rm 2+}{\it n}}{\rm (}{\rm 1}+{\it n}{\rm )!}}}+{\rm (}{\rm 2}{\it n}+{\rm 2}{\rm )(}{\rm 2}{\rm )}{\it a}_{{\rm 2}{\it n}{\rm +2}}+{\it a}_{{\rm 2}{\it n}}={\rm 0}$$ $$\Rightarrow-{{{\it n}+{\rm 2}}\over{{\rm 2}^{{\it n}{\rm +1}}{\rm (}{\it n}+{\rm 1}{\rm )!}}}+{\rm 2}{\rm (}{\rm 2}{\it n}+{\rm 2}{\rm )}{\it a}_{{\rm 2}{\it n}{\rm +2}}-{{{\rm 1}+{{{\rm 1}}\over{{\rm 2}}}+\cdots+{{{\rm 1}}\over{{\it n}}}}\over{{\rm 2}^{{\it n}{\rm +1}}{\it n}{\rm !}}}={\rm 0}$$ $$\Rightarrow{\rm 2}^{{\rm 2}}{\rm (}{\it n}+{\rm 1}{\rm )}{\it a}_{{\rm 2}{\it n}{\rm +2}}={{{\it n}+{\rm 2}}\over{{\rm 2}^{{\it n}{\rm +1}}{\rm (}{\it n}+{\rm 1}{\rm )!}}}+{{{\rm 1}+{{{\rm 1}}\over{{\rm 2}}}+\cdots+{{{\rm 1}}\over{{\it n}}}}\over{{\rm 2}^{{\it n}{\rm +1}}{\it n}{\rm !}}}={{{\it n}+{\rm 2}+{\rm (}{\it n}+{\rm 1}{\rm )(}{\rm 1}+{{{\rm 1}}\over{{\rm 2}}}+\cdots+{{{\rm 1}}\over{{\it n}}}{\rm )}}\over{{\rm 2}^{{\it n}{\rm +1}}{\rm (}{\it n}+{\rm 1}{\rm )!}}}$$ $$\Rightarrow{\it a}_{{\rm 2}{\it n}{\rm +2}}={{{\rm 1}+{{{\rm 1}}\over{{\it n}+{\rm 1}}}+{\rm 1}+{{{\rm 1}}\over{{\rm 2}}}+\cdots+{{{\rm 1}}\over{{\it n}}}}\over{{\rm 2}^{{\it n}{\rm +3}}{\rm (}{\it n}+{\rm 1}{\rm )!}}}$$ For the coefficient of $x^{4+n}$: $${\rm 2}{{{\rm 4}+{\it n}}\over{{\rm 2}^{{\rm 4+}{\it n}}{\rm 2}{\rm !(}{\rm 2}+{\it n}{\rm )!}}}+{\rm (}{\rm 2}{\it n}+{\rm 4}{\rm )(}{\rm 4}{\rm )}{\it a}_{{\rm 2}{\it n}{\rm +4}}+{\it a}_{{\rm 2}{\it n}{\rm +2}}={\rm 0}$$ $${{{\it n}+{\rm 4}}\over{{\rm 2}^{{\it n}{\rm +3}}{\rm 2}{\rm !(}{\it n}+{\rm 2}{\rm )!}}}+{\rm 2}^{{\rm 3}}{\rm (}{\it n}+{\rm 2}{\rm )}{\it a}_{{\rm 2}{\it n}{\rm +4}}+{{{\rm 1}+{\rm (}{\rm 1}+{{{\rm 1}}\over{{\rm 2}}}+\cdots+{{{\rm 1}}\over{{\it n}+{\rm 1}}}{\rm )}}\over{{\rm 2}^{{\it n}{\rm +3}}{\rm (}{\it n}+{\rm 1}{\rm )!}}}={\rm 0}$$ $${\rm 2}^{{\rm 3}}{\rm (}{\it n}+{\rm 2}{\rm )}{\it a}_{{\rm 2}{\it n}{\rm +4}}=-{{{\it n}+{\rm 4}}\over{{\rm 2}^{{\it n}{\rm +3}}{\rm 2}{\rm !(}{\it n}+{\rm 2}{\rm )!}}}-{{{\rm 1}+{\rm (}{\rm 1}+{{{\rm 1}}\over{{\rm 2}}}+\cdots+{{{\rm 1}}\over{{\it n}+{\rm 1}}}{\rm )}}\over{{\rm 2}^{{\it n}{\rm +3}}{\rm (}{\it n}+{\rm 1}{\rm )!}}}=-{{{\it n}+{\rm 4}+{\rm 2}{\rm (}{\it n}+{\rm 2}{\rm )}\left[{{\rm 1}+{\rm (}{\rm 1}+{{{\rm 1}}\over{{\rm 2}}}+\cdots+{{{\rm 1}}\over{{\it n}+{\rm 1}}}{\rm )}}\right]}\over{{\rm 2}^{{\it n}{\rm +3}}{\rm 2}{\rm !(}{\it n}+{\rm 2}{\rm )!}}}$$ $${\it \Rightarrow}{\it a}_{{\rm 2}{\it n}{\rm +4}}=-{{{{{\it n}+{\rm 4}}\over{{\rm 2}{\rm (}{\it n}+{\rm 2}{\rm )}}}+{\rm 1}+{\rm (}{\rm 1}+{{{\rm 1}}\over{{\rm 2}}}+\cdots+{{{\rm 1}}\over{{\it n}+{\rm 1}}}{\rm )}}\over{{\rm 2}^{{\it n}{\rm +5}}{\rm 2}{\rm !(}{\it n}+{\rm 2}{\rm )!}}}=-{{{{{\rm 1}}\over{{\rm 2}}}+{{{\rm 1}}\over{{\it n}+{\rm 2}}}+{\rm 1}+{\rm (}{\rm 1}+{{{\rm 1}}\over{{\rm 2}}}+\cdots+{{{\rm 1}}\over{{\it n}+{\rm 1}}}{\rm )}}\over{{\rm 2}^{{\it n}{\rm +5}}{\rm 2}{\rm !(}{\it n}+{\rm 2}{\rm )!}}}$$ $$=-{{{\rm (}{\rm 1}+{{{\rm 1}}\over{{\rm 2}}}{\rm )}+{\rm (}{\rm 1}+{{{\rm 1}}\over{{\rm 2}}}+\cdots+{{{\rm 1}}\over{{\it n}+{\rm 1}}}+{{{\rm 1}}\over{{\it n}+{\rm 2}}}{\rm )}}\over{{\rm 2}^{{\it n}{\rm +5}}{\rm 2}{\rm !(}{\it n}+{\rm 2}{\rm )!}}}$$ And in general, it can be shown by induction that: $${\it a}_{{\rm 2}{\it m}}={\rm (}-{\rm 1}{\rm )}^{{\it m}{\rm -}{\it n}{\rm +1}}{{{\rm (}{\rm 1}+{{{\rm 1}}\over{{\rm 2}}}+\cdots+{{{\rm 1}}\over{{\it m}-{\it n}}}{\rm )}+{\rm (}{\rm 1}+{{{\rm 1}}\over{{\rm 2}}}+\cdots+{{{\rm 1}}\over{{\it m}}}{\rm )}}\over{{\rm 2}^{{\rm 2}{\it m}{\rm -}{\it n}{\rm +1}}{\rm (}{\it m}-{\it n}{\rm )!}{\it m}{\rm !}}}{\rm }{\rm (}{\it m}\geq{\it n}{\rm )}{\rm (}II{\rm )}$$ Therefore, according to relations ${\rm (}I )$ and ${\rm (}II )$ $${\it y}_{{\rm 2}}{\rm (}{\it x}{\rm )}={\it J}_{{\it n}}{\rm (}{\it x}{\rm )}{\it Lnx}+\sum\limits_{{\it m}{\rm =0}}^{{\it n}{\rm -1}}{-{{{\rm (}{\it n}-{\it m}-{\rm 1}{\rm )!}}\over{{\rm 2}^{{\rm 2}{\it m}{\rm -}{\it n}{\rm +1}}{\it m}{\rm !}}}}{\it x}^{{\rm 2}{\it m}{\rm -}{\it n}}+\sum\limits_{{\it m}={\it n}}^{\infty}{{\rm (}-{\rm 1}{\rm )}^{{\it m}{\rm -}{\it n}{\rm +1}}}{{{\rm (}{\rm 1}+\cdots+{{{\rm 1}}\over{{\it m}-{\it n}}}{\rm )}+{\rm (}{\rm 1}+\cdots+{{{\rm 1}}\over{{\it m}}}{\rm )}}\over{{\rm 2}^{{\rm 2}{\it m}{\rm -}{\it n}{\rm +1}}{\rm (}{\it m}-{\it n}{\rm )!}{\it m}{\rm !}}}{\it x}^{{\rm 2}{\it m}{\rm -}{\it n}}$$ $$={\it J}_{{\it n}}{\rm (}{\it x}{\rm )}{\it Lnx}-{{{\rm 1}}\over{{\rm 2}}}\sum\limits_{{\it m}{\rm =0}}^{{\it n}{\rm -1}}{{{{\rm (}{\it n}-{\it m}-{\rm 1}{\rm )!}}\over{{\it m}{\rm !}}}}{{{\it x}^{{\rm 2}{\it m}{\rm -}{\it n}}}\over{{\rm 2}^{{\rm 2}{\it m}{\rm -}{\it n}}}}-{{{\rm 1}}\over{{\rm 2}}}\sum\limits_{{\it m}={\it n}}^{\infty}{{\rm (}-{\rm 1}{\rm )}^{{\it m}{\rm -}{\it n}}}{{{\rm (}{\rm 1}+\cdots+{{{\rm 1}}\over{{\it m}-{\it n}}}{\rm )}+{\rm (}{\rm 1}+\cdots+{{{\rm 1}}\over{{\it m}}}{\rm )}}\over{{\rm (}{\it m}-{\it n}{\rm )!}{\it m}{\rm !}}}{{{\it x}^{{\rm 2}{\it m}{\rm -}{\it n}}}\over{{\rm 2}^{{\rm 2}{\it m}{\rm -}{\it n}}}}$$ $$={\it J}_{{\it n}}{\rm (}{\it x}{\rm )}{\it Lnx}-{{{\rm 1}}\over{{\rm 2}}}\sum\limits_{{\it m}{\rm =0}}^{{\it n}{\rm -1}}{{{{\rm (}{\it n}-{\it m}-{\rm 1}{\rm )!}}\over{{\it m}{\rm !}}}}{\rm (}{{{\it x}}\over{{\rm 2}}}{\rm )}^{{\rm 2}{\it m}{\rm -}{\it n}}-{{{\rm 1}}\over{{\rm 2}}}\sum\limits_{{\it m}={\it n}}^{\infty}{{\rm (}-{\rm 1}{\rm )}^{{\it m}{\rm -}{\it n}}}{{{\rm (}{\rm 1}+\cdots+{{{\rm 1}}\over{{\it m}-{\it n}}}{\rm )}+{\rm (}{\rm 1}+\cdots+{{{\rm 1}}\over{{\it m}}}{\rm )}}\over{{\rm (}{\it m}-{\it n}{\rm )!}{\it m}{\rm !}}}{\rm (}{{{\it x}}\over{{\rm 2}}}{\rm )}^{{\rm 2}{\it m}{\rm -}{\it n}}$$ If we reduce the lower limit of the last sigma by $n$ units, we have: $${\it y}_{{\rm 2}}{\rm (}{\it x}{\rm )}={\it J}_{{\it n}}{\rm (}{\it x}{\rm )}{\it Lnx}-{{{\rm 1}}\over{{\rm 2}}}\sum\limits_{{\it m}{\rm =0}}^{{\it n}{\rm -1}}{{{{\rm (}{\it n}-{\it m}-{\rm 1}{\rm )!}}\over{{\it m}{\rm !}}}}{\rm (}{{{\it x}}\over{{\rm 2}}}{\rm )}^{{\rm 2}{\it m}{\rm -}{\it n}}-{{{\rm 1}}\over{{\rm 2}}}\sum\limits_{{\it m}={\rm 0}}^{\infty}{{\rm (}-{\rm 1}{\rm )}^{{\it m}}}{{{\rm (}{\rm 1}+\cdots+{{{\rm 1}}\over{{\it m}}}{\rm )}+{\rm (}{\rm 1}+\cdots+{{{\rm 1}}\over{{\it m}+{\it n}}}{\rm )}}\over{{\it m}{\rm !(}{\it m}+{\it n}{\rm )!}}}{\rm (}{{{\it x}}\over{{\rm 2}}}{\rm )}^{{\rm 2}{\it m}{\rm +}{\it n}}$$ Then, instead of the solution ${\it y}_{{\rm 2}}$, we consider the following solution which is linearly independent with ${\it J}_{{\it n}}{\rm (}{\it x}{\rm )}$: $${\it Y}_{{\it n}}{\rm (}{\it x}{\rm )}={\it a}{\rm (}{\it y}_{{\rm 2}}{\rm (}{\it x}{\rm )}+{\it bJ}_{{\it n}}{\rm (}{\it x}{\rm ))}$$ where ${\it a}={{{\rm 2}}\over{{\rm \pi}}}$ and ${\it b}={\rm \gamma}-{\it Ln}{\rm 2}$. So the solution is $${\it Y}_{{\it n}}{\rm (}{\it x}{\rm )}={{{\rm 2}}\over{{\rm \pi}}}\left({{\it y}_{{\rm 2}}{\rm (}{\it x}{\rm )}+{\rm (}{\rm \gamma}-{\it Ln}{\rm 2}{\rm )}{\it J}_{{\it n}}{\rm (}{\it x}{\rm )}}\right)$$ $${\it Y}_{{\it n}}{\rm (}{\it x}{\rm )}={{{\rm 2}}\over{{\rm \pi}}}\left({{\rm (}{\it Lnx}+{\rm \gamma}-{\it Ln}{\rm 2}{\rm )}{\it J}_{{\it n}}{\rm (}{\it x}{\rm )}-{{{\rm 1}}\over{{\rm 2}}}\sum\limits_{{\it m}{\rm =0}}^{{\it n}{\rm -1}}{{{{\rm (}{\it n}-{\it m}-{\rm 1}{\rm )!}}\over{{\it m}{\rm !}}}}{\rm (}{{{\it x}}\over{{\rm 2}}}{\rm )}^{{\rm 2}{\it m}{\rm -}{\it n}}-{{{\rm 1}}\over{{\rm 2}}}\sum\limits_{{\it m}={\rm 0}}^{\infty}{{\rm (}-{\rm 1}{\rm )}^{{\it m}}}{{{\rm (}{\rm 1}+\cdots+{{{\rm 1}}\over{{\it m}}}{\rm )}+{\rm (}{\rm 1}+\cdots+{{{\rm 1}}\over{{\it m}+{\it n}}}{\rm )}}\over{{\it m}{\rm !(}{\it m}+{\it n}{\rm )!}}}{\rm (}{{{\it x}}\over{{\rm 2}}}{\rm )}^{{\rm 2}{\it m}{\rm +}{\it n}}}\right)$$ $$={{{\rm 2}}\over{{\rm \pi}}}{\rm (}{\it Lnx}+{\rm \gamma}-{\it Ln}{\rm 2}{\rm )}{\it J}_{{\it n}}{\rm (}{\it x}{\rm )}-{{{\rm 1}}\over{{\rm \pi}}}\sum\limits_{{\it m}{\rm =0}}^{{\it n}{\rm -1}}{{{{\rm (}{\it n}-{\it m}-{\rm 1}{\rm )!}}\over{{\it m}{\rm !}}}}{\rm (}{{{\it x}}\over{{\rm 2}}}{\rm )}^{{\rm 2}{\it m}{\rm -}{\it n}}-{{{\rm 1}}\over{{\rm \pi}}}\sum\limits_{{\it m}={\rm 0}}^{\infty}{{\rm (}-{\rm 1}{\rm )}^{{\it m}}}{{{\rm (}{\rm 1}+\cdots+{{{\rm 1}}\over{{\it m}}}{\rm )}+{\rm (}{\rm 1}+\cdots+{{{\rm 1}}\over{{\it m}+{\it n}}}{\rm )}}\over{{\it m}{\rm !(}{\it m}+{\it n}{\rm )!}}}{\rm (}{{{\it x}}\over{{\rm 2}}}{\rm )}^{{\rm 2}{\it m}{\rm +}{\it n}}$$.
One way is to expand term by term in the solution $$ \text{Y}_{n}(x) = \text{J}_{n}(x) \, \int \frac{dx}{x \, \text{J}_{n}^{2}(x)}. $$ Another way is to use the ratio $$ \text{Y}_{\nu}(x) = \frac{\cos(\nu \pi) \, \text{J}_{\nu}(x) - \text{J}_{-\nu}(x)}{\sin(\nu \pi)} $$ as $ \nu \to n$ with $n$ being an integer. This can also be stated as $$\text{Y}_{n}(x) = \left[ \frac{\partial \, \text{J}_{\nu}(x)}{\partial \nu} - (-1)^n \, \frac{\partial \, \text{J}_{-\nu}(x)}{\partial \nu} \right]_{\nu = n} $$ and leads to $$ \text{Y}_{0}(x) = 2 \, \left[ \frac{\partial \, \text{J}_{\nu}(x)}{\partial \nu} \right]_{\nu = 0}. $$
See G. N. Watson, A Treatise on the theory of Bessel Functions starting around page 67.
The derivatives can be seen to operate as seen in this example: Consider $$ \frac{\partial \, J_{\nu}}{\partial \nu} = \sum_{k=0}^{\infty} \frac{(-1)^k \, \left(\frac{x}{2}\right)^{2 k}}{k!} \, \frac{\partial}{\partial \nu} \, \frac{\left(\frac{x}{2}\right)^{\nu}}{\Gamma(k+\nu+1)}. $$ Using $t^{\nu} = e^{\nu \, \ln(t)}$ then $$ \frac{\partial}{\partial \nu} \, \frac{\left(\frac{x}{2}\right)^{\nu}}{\Gamma(k+\nu+1)} = \frac{\left(\frac{x}{2}\right)^{\nu}}{\Gamma(k+\nu+1)} \, \left( \ln\left(\frac{x}{2}\right) - \psi(k+\nu+1) \right), $$ where $\psi(x)$ is the digmamma function, leads to $$ \frac{\partial \, J_{\nu}}{\partial \nu} = \sum_{k=0}^{\infty} \frac{(-1)^k \, \left(\frac{x}{2}\right)^{2 k + \nu}}{k! \, (k+\nu)!} \, \left( \ln\left(\frac{x}{2}\right) - \psi(k+\nu+1) \right) $$ for which \begin{align} \text{Y}_{0}(x) &= 2 \, \left[ \frac{\partial \, \text{J}_{\nu}(x)}{\partial \nu} \right]_{\nu = 0} \\ &= 2 \, \left. \sum_{k=0}^{\infty} \frac{(-1)^k \, \left(\frac{x}{2}\right)^{2 k + \nu}}{k! \, (k+\nu)!} \, \left( \ln\left(\frac{x}{2}\right) - \psi(k+\nu+1) \right) \right|_{\nu = 0} \end{align} or \begin{align} \text{Y}_{0}(x) &= 2 \, \sum_{k=0}^{\infty} \frac{(-1)^k \, \left(\frac{x}{2}\right)^{2 k}}{(k!)^2} \, \left( \ln\left(\frac{x}{2}\right) - \psi(k+1) \right) \\ &= 2 \, \left( \text{J}_{0}(x) \, \ln\left(\frac{x}{2}\right) - \sum_{k=0}^{\infty} \frac{(-1)^{k} \, \left(\frac{x}{2}\right)^{2k}}{(k!)^2} \, \psi(k+1) \right) \\ &= 2 \, \left(\gamma + \ln\left(\frac{x}{2}\right)\right) \, \text{J}_{0}(x) - 2 \, \sum_{k=1}^{\infty} \frac{(-1)^{k} \, \left(\frac{x}{2}\right)^{2k}}{(k!)^2} \, H_{k}, \end{align} where $H_{n}$ is the harmonic number. These last three expressions are related by making use of the digamma function. Working out the details for the general case follows in a similar manor as the example.