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 me rewrite your ODE as follows:

$$L[y] = p\_0 y'' + p\_1 y' + p\_2 y = 0,$$ where $p\_i(x)$ are the coefficients of the equation. If you know one of the two linear indepent solutions of the homogenous part (indeed the whole ode is homogenouse), say $y\_1$, you can obtain $y\_2$ ($y = A y\_1+B y\_2$) by using the method of *variation of parameters*.

Let me show you.

Set $y = A(x)y\_1$ and substitute back into the original ode:

$$p\_0 (A''y\_1 + 2A'y\_1' + A y\_1'') + p\_1 (A'y\_1 + A y\_1') + p\_2 A y\_1 = 0,$$

arrange terms in order to get a ODE-1 for $A(x)$:

$$ A'' + \left(\frac{2y'\_1}{y\_1} + \frac{p\_1 y\_1}{p\_0} \right) A' + \frac{1}{p\_0} L[y\_1] A= 0,$$

since $L[y\_1]=0$, $A$ can be obtained by solving this equation in terms of an integrating factor as follows:

$$\frac{d}{dx} (u A') = u \cdot 0 = 0,$$

where $u = e^{\int \left(\frac{2y'\_1}{y\_1} + \frac{p\_1 y\_1}{p\_0} \right) \, dx}$. It yields:

$$uA' = a\_2 \Rightarrow A = a\_1 + a\_2 \int \frac{dx}{u},$$

and hence the solution, $y = a\_1 y\_1 + a\_2 y\_1 \int \frac{dx}{u}$, being $a\_i$ constants of integration.

I hope this is useful to you.

Cheers!