<s>[INST] Solving a 2nd order differential equation by the Frobenius method Can you, please, help me to solve this equation:
$$(x+1)^2y''+(x+1)y'-y=0$$
Here, for me the problem is, I am finding relationship among 3 members: $a_n, a_{n+1}, a_{n+2}$, not between 2 members: $a_n$ and $a_{n+1}$
I would like to solve it using the Frobenius method.
 [/INST] Let $y=\sum\limits_{n=0}^\infty a_n(x+1)^{n+r}$ ,
Then $y'=\sum\limits_{n=0}^\infty(n+r)a_n(x+1)^{n+r-1}$
$y''=\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_n(x+1)^{n+r-2}$
$\therefore(x+1)^2\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_n(x+1)^{n+r-2}+(x+1)\sum\limits_{n=0}^\infty(n+r)a_n(x+1)^{n+r-1}-\sum\limits_{n=0}^\infty a_n(x+1)^{n+r}=0$
$\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_n(x+1)^{n+r}+\sum\limits_{n=0}^\infty(n+r)a_n(x+1)^{n+r}-\sum\limits_{n=0}^\infty a_n(x+1)^{n+r}=0$
$\sum\limits_{n=0}^\infty((n+r)^2-1)a_n(x+1)^{n+r}=0$
Since there are not any indical equations present, that means $r$ can be chosen as any complex number.
However, in fact, take $r=1$ or $r=-1$ will bring the above equation most simplified.
Moreover, in fact, we can find all groups of the linearly independent solutions by just taking $r=-1$ :
$\sum\limits_{n=0}^\infty((n-1)^2-1)a_n(x+1)^{n-1}=0$
$\sum\limits_{n=0}^\infty n(n-2)a_n(x+1)^{n-1}=0$
$\therefore n(n-2)a_n=0$
$\therefore\begin{cases}a_0=a_0\\a_2=a_2\\a_n=0~\forall n\in\mathbb{N}\setminus\{2\}\end{cases}$
Hence $y=\dfrac{C_1}{x+1}+C_2(x+1)$
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