Determining the first $3$ nonzero terms in each of two linearly independent solutions In preparation for an upcoming exam I have came across a question that I am a little confused with.
Given the following different equation,
$$xy''+y'-y = 0$$
I am trying to find the first $3$ nonzero terms in each of two linearly independent solutions.
I have determined the indicial equation to be,
$$r^2 = 0$$
Therefore the roots are,
$$r_1=0, r_2=0$$
So the roots are separated by a integer
I have calculated the recurrence solution to be,
$$a_n = \frac{a_{n-1}}{(n+r)^2}$$
So therefore the first solution would be the following,
$$y_1(x) = 1+ x + \frac{1}{4}x^2 + \frac{1}{36}x^3...$$
However it is determining the second solution where I am having problem, so I am looking for some help with showing how this answer is calculated, thanks! The solution to the second solution is,
$$y_2(x) = y_1(x)\ln x -2x - \frac{3}{4}x^2-\frac{11}{108}x^3$$

Since $r_1=r_2=0$, a second solution $y_2(x)$ has the form
$$
y_2(x) = y_1(x)\ln x +\sum_{n=1}^{\infty}b_nx^n.
$$
We have
$$
y_2'(x)=y_1'\ln(x)+y_1x^{-1}+\sum_{n=1}^{\infty}nb_nx^{n-1}\\
y_2''(x)=y_1''\ln(x)+2x^{-1}y_1'-y_1x^{-2}+\sum_{n=2}^{\infty}n(n-1)b_nx^{n-2}.
$$
Inserting this in the equation we get
$$
2y_1'+\sum_{n=2}^{\infty}n(n-1)b_nx^{n-1}+\sum_{n=1}^{\infty}nb_nx^{n-1}-\sum_{n=1}^{\infty}b_nx^n=0,
$$
or equivalently, we write
$$
2a_1+2\sum_{n=2}^{\infty}na_nx^{n-1}+\sum_{n=2}^{\infty}n(n-1)b_nx^{n-1}+b_1+\sum_{n=2}^{\infty}nb_nx^{n-1}-\sum_{n=2}^{\infty}b_{n-1}x^{n-1}=0.
$$
Thus
$$2a_1+b_1+\sum_{n=2}^{\infty}(2na_n+n^2n_n-b_{n-1})x^{n-1}=0.$$
By identification, we obtain
$$
\left\{\begin{array}{l}
b_1=-2a_1\\
b_n=\frac{b_{n-1}-2na_n}{n^2},\quad n\geq2. \end{array}\right. \qquad\Rightarrow\left\{\begin{array}{l}
b_1=-2\\
b_2=-\dfrac{3}{4}\\
b_3=-\dfrac{11}{108}  \end{array}\right.
$$
