Consider the differential equation
$$x^2y''+3(x-x^2)y'-3y=0$$

$(a)$ Find the recurrence equation and first three nonzero terms of the series solution in powers of $$ corresponding to the larger root of the indicial equation.

$(b)$ What would be the form of a second linearly independent solution of this differential equation?

I found the indicial equation to be $r(r-1)+3r-3 = 0$, so the two roots are $r\_1=-3$ and $r\_2=1$.

And the recurrence relation is $a\_n = \dfrac{3(n-1+r)a\_{n-1}}{(n+r)(n+r-1)+3(n+r)-3}$

set $r=1$, then $a\_n = \dfrac{3na\_{n-1}}{(n+3)(n+1)-3}$. And I can figure out first linearly independent solution.

But for $r=-3$,then $a\_n = \dfrac{3(n-4)a\_{n-1}}{(n-3)(n-1)-3}$, if let $a\_0$ be an arbitrary constant, then $a\_1 =\dfrac{3(-3)a\_0}{(-2)(0)}$, which doesn't work.

Then how do I figure out the second linearly independent solution?
['https://math.stackexchange.com/questions/652362', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/59036/']
Here is the procedure.

Denote the two roots by $r\_1$ and $r\_2$, with $r\_1 \gt r\_2$.

The Method of Frobenius will always generate a solution corresponding to $r\_1$, but **may** generate a solution for the smaller second root $r\_2$ of the indicial equation.

If the method fails for $r\_2$, then an approach is to keep the recursion solution in terms of $r$ and use it to find the coefficients $a\_n$ (for $n \ge 1$), in terms of both $r$ and $a\_0$, where $a\_0 \ne 0$. For ease, $a\_0$ is typically chose to be one.

Using this more general form and the coefficients, the two independent solutions can be written as:

$$y\_1(r, x) = x^r \sum\_{n=0}^\infty ~ a\_n(r)x^n = \sum\_{n=0}^\infty ~ a\_n(r)x^{n+r} \\y\_2(r,x) = \dfrac{\partial}{\partial r}[(r - r\_2)y\_1(r, x)]~\Bigr|\_{r=r\_2}$$

You should be able to use this approach and show:

$$y(x) = y\_1(x) + y\_2(x) = \dfrac{c\_1(3x(3x^2 + 3x+ 2) + 2)}{x^3} + \dfrac{c\_2e^{3x}}{x^3}$$
Let $y=\sum\limits\_{n=0}^\infty a\_nx^{n+r}$ ,

Then $y'=\sum\limits\_{n=0}^\infty(n+r)a\_nx^{n+r-1}$

$y''=\sum\limits\_{n=0}^\infty(n+r)(n+r-1)a\_nx^{n+r-2}$

$\therefore x^2\sum\limits\_{n=0}^\infty(n+r)(n+r-1)a\_nx^{n+r-2}+3(x-x^2)\sum\limits\_{n=0}^\infty(n+r)a\_nx^{n+r-1}-3\sum\limits\_{n=0}^\infty a\_nx^{n+r}=0$

$\sum\limits\_{n=0}^\infty(n+r)(n+r-1)a\_nx^{n+r}+\sum\limits\_{n=0}^\infty3(n+r)a\_nx^{n+r}-\sum\limits\_{n=0}^\infty3(n+r)a\_nx^{n+r+1}-\sum\limits\_{n=0}^\infty 3a\_nx^{n+r}=0$

$\sum\limits\_{n=0}^\infty(n+r+3)(n+r-1)a\_nx^{n+r}-\sum\limits\_{n=1}^\infty3(n+r-1)a\_{n-1}x^{n+r}=0$

$(r+3)(r-1)a\_0x^r+\sum\limits\_{n=1}^\infty((n+r+3)(n+r-1)a\_n-3(n+r-1)a\_{n-1})x^{n+r}=0$

The indicial equation is $(r+3)(r-1)=0$ , the two roots are $r\_1=-3$ and $r\_2=1$ .

The recurrence relation is $(n+r+3)(n+r-1)a\_n-3(n+r-1)a\_{n-1}=0$ , i.e. $(n+r+3)(n+r-1)a\_n=3(n+r-1)a\_{n-1}$

Each root from the indical equations may not always only find one group of the linearly independent solutions, sometimes we can find more than one group of the linearly independent solutions at the same time.

When we choose $r=-3$ , the recurrence relation becomes $n(n-4)a\_n=3(n-4)a\_{n-1}$ , which has a value of $n$ at $n=4$ so that it is independent to the recurrence relation. Besides $a\_0$ can chooes arbitrary, $a\_4$ can also chooes arbitrary. This makes the effect that we can find two groups of the linearly independent solutions that basing on $n=0$ to $n=3$ and $n=4$ to $n=+\infty$ respectively.