Reduction of a Power Series Solution Separated by an Integer? I am given the following equation: $$x(x-1)\ddot y + 6x^2\dot y + 3y = 0 $$
and have solved/found the regular singular points, the exponents at singularity, the first series solution and the form of the second series solution, which are all as follows:
$x = 0, \space 1$ are regular singular points
For $x = 0$, $r_1 = 1, \space r_2 = 0$
Recurrence relationship: $$a_n = \frac {n^2 - n + 3}{n(n+1)}a_{n-1} + \frac {6(n-1)}{n(n+1)}a_{n-2}$$
$a_1 = \frac{3}{2}a_0$
The first solution is:
$$y_1(x) = x + \frac{3}{2}x^2 + \frac{9}{4}x^3 + \frac{51}{16}x^4 + ...$$
Since the exponents differ by an integer, the second solution is of the form:
$$y_2(x) = ay_1(x)ln(x) + 1 + \sum_{n=1}^{\infty} c_n(r_2)x^n$$
Substituting this equation into the original governing equation:
$$x(x-1)\bigg\{ay_1(x)ln(x) + 1 + \sum_{n=1}^{\infty} c_n(r_2)x^n)\bigg\}^" + 6x^2 \bigg\{ ay_1(x)ln(x) + 1 + \sum_{n=1}^{\infty} c_n(r_2)x^n \bigg\} ' + 3\bigg\{ ay_1(x)ln(x) + 1 + \sum_{n=1}^{\infty} c_n(r_2)x^n\bigg\} = 0 $$
Now my question is how does the above equation reduce to the following?:
$$aln(x)L[y_1] + 2a(x-1)y' - \frac{(x-1)}{x}ay_1 + 6axy_1 + L\bigg[ 1 + \sum_{n=1}^{\infty} c_n(r_2)x^n \bigg] = 0$$
The above form is advantageous since $L[y_1] = 0$, but I do not fully understand this reduction step.
Thanks!

Just take derivative and separate:
Terms that $\ln$ has no derivative in them, they produce first term.
Terms that $\ln$ has derivative in them, make middle terms.
Terms that summation has derivative make last term.
