The second solution for $xy''+y=0$. 
Solve $xy''+y=0$. 

After obtaining the first solution $y_1(x)$ using the Frobenius series
$$y_1(x) = a_0\sum_{n=0}^\infty{\frac{(-1)^n}{n!(n+1)!}x^{n+1}}$$
I need to find the second solution which will be of the form 
$$y_1(x)\ln(x)+C(x), \qquad C(x) = \sum_{n=0}^\infty{c_nx^n}$$ 
but I'm having problems with all the indices. Thanks in advance.
 A: Let $y(x)=\sum_{n=0}^\infty c_nx^{n+r}$ be a solution.  Then $$y''(x)=\sum_{n=0}^\infty (n+r)(n+r-1)c_nx^{n+r-2}.$$
Therefore
$$xy''(x)=\sum_{n=0}^\infty(n+r)(n+r-1)c_nx^{n+r-1}.$$
Hence
$$xy''(x)+y(x)=r(r-1)c_0x^{r-1}+\sum_{n=0}^\infty (n+r)(n+r+1)c_{n+1}x^{n+r}+\sum_{n=0}^\infty c_nx^{n+r}.$$
This requires $r(r-1)=0$ and
$$(n+r)(n+r+1)c_{n+1}=-c_n\tag{1}$$
for all $n\ge 0$.
The roots $r=0$ and $r=1$ of $r(r-1)=0$ differ by an integer.  Therefore, there exist two linearly independent solutions
$$y_1(x)=\sum_{n=0}^\infty c_n x^{n+1}$$
and
$$y_2(x)=ky_1(x)\ln x+\sum_{n=0}^\infty b_n x^{n+0}.$$
If $c_0=1$, then from $(1)$ with $r=1$, we conclude that $$c_n=\frac{(-1)^n}{n!(n+1)!}$$ for all $n\ge 0$.   That is,
$$y_1(x)=\sum_{n=0}^\infty\frac{(-1)^n}{n!(n+1)!}x^{n+1}.$$
We now want to find $y_2(x)$.  Up to rescaling, we may assume $k=1$.    Note that
$$y_2'(x)=\frac{y_1(x)}{x}+y'_1(x)\ln x+\sum_{n=0}^\infty nb_nx^{n-1}$$
and
$$y_2''(x)=-\frac{y_1(x)}{x^2}+\frac{2y'_1(x)}{x}+y_1''(x)\ln x+\sum_{n=0}^\infty n(n-1)b_nx^{n-2}.$$
Hence
$$0=xy_2''(x)+y_2(x)=-\frac{y_1(x)}{x}+2y'_1(x)+xy_1''(x)\ln x+\sum_{n=0}^\infty n(n-1) b_nx^{n-1}+y_1(x)\ln x+\sum_{n=0}^\infty b_n x^n.$$
But $xy_1''(x)+y_1(x)=0$, so
$$0=xy_2''(x)+y_2(x)=-\frac{y_1(x)}{x}+2y_1'(x)+\sum_{n=0}^\infty \big((n+1)nb_{n+1}+b_n\big)x^n.$$
Note that
$$\frac{y_1(x)}{x}=\sum_{n=0}^\infty\frac{(-1)^n}{n!(n+1)!}x^n$$
and
$$y_1'(x)=\sum_{n=0}^\infty\frac{(-1)^n}{n!n!}x^n.$$
Therefore we have
$$\sum_{n=0}^\infty\left((n+1)nb_{n+1}+b_n-\frac{(-1)^n}{n!(n+1)!}+\frac{2(-1)^n}{n!n!}\right)x^n=0.$$
Thus
$$(n+1)nb_{n+1}+b_n=\frac{(-1)^{n+1}(2n+1)}{n!(n+1)!}$$
for all $n=0,1,2,\ldots$.  This shows that $b_0=-1$. 
For $n\ge 2$, 
$$b_{n}=\frac{(-1)^{n}(2n-1)}{(n-1)n!n!}-\frac{b_{n-1}}{n(n-1)}.$$
Let $B_n=(-1)^nn!(n-1)!b_n$, so $B_n=\frac{2n-1}{n(n-1)}+B_{n-1}$.
Thus
$$B_n-B_{n-1}=\frac{1}{n-1}+\frac{1}{n}$$
for $n\ge 2$.  Hence
$$B_n-B_1=\left(1+\frac12\right)+\left(\frac12+\frac13\right)+\ldots+\left(\frac1{n-1}+\frac1n\right).$$
Wlog, we can set $B_1=1$ (otherwise we add an appropriate multiple of $y_1(x)$ to $y_2(x)$).  Therefore,
$$B_n=2H_n-\frac1n,$$
where $H_n=\sum_{j=1}^n\frac1j$ is the $n$th Harmonic number (with $H_0=0$).  This shows that
$$b_n=\frac{(-1)^n}{n!n!}\left(2nH_n-1\right).$$
Observe that this formula works for $n=0$ as well.  That is
$$y_2(x)=\sum_{n=0}^\infty \frac{(-1)^n}{n!(n+1)!}x^{n+1}\ln x+\sum_{n=0}^\infty  \frac{(-1)^n}{n!n!}\left(2nH_n-1\right)x^n.$$
All solutions $y(x)$ to $xy''(x)+y(x)=0$ are linear combinations of $y_1(x)$ and $y_2(x)$.
