Question

Find the first three nonzero terms in each of two solutions (noy1t multiples of each other) xy''+y'-y=0

Answer 1

It looks like you're trying to solve the ODE via Frobenius' method. It's not immediately clear where the solution is to be centered, but I think we can safely assume it's to be taken about $x=0$. Dividing through by $x$, we get

$y''+\dfrac{y'}x-\dfrac yx=0$

and we observe that $x=0$ is a regular singular point.

Now, taking a solution of the form

$y=\displaystyle\sum_{n\ge0}a_nx^{n+r}$
$\implies y'=\displaystyle\sum_{n\ge0}(n+r)a_nx^{n+r-1}$$\implies y''=\displaystyle\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}$

and substituting into the ODE gives

$\displaystyle\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-1}+\sum_{n\ge0}(n+r)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r}=0$
$\displaystyle\sum_{n\ge0}(n+r)^2a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r}=0$
$\displaystyle r^2a_0x^{r-1}+\sum_{n\ge1}(n+r)^2a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r}=0$
$\displaystyle r^2a_0x^{r-1}+\sum_{n\ge1}(n+r)^2a_nx^{n+r-1}-\sum_{n\ge1}a_{n-1}x^{n+r-1}=0$
$\displaystyle r^2a_0x^{r-1}+\sum_{n\ge1}\bigg((n+r)^2a_n-a_{n-1}\bigg)x^{n+r-1}=0$

The indicial polynomial admits only one root (of multiplicity two) at $r=0$, so in fact a regular power series solution will exist. So in fact the ODE above reduces to

$\displaystyle\sum_{n\ge1}\bigg(n^2a_n-a_{n-1}\bigg)x^{n-1}=\sum_{n\ge0}\bigg((n+1)^2a_{n+1}-a_n\bigg)x^n=0$

leaving us with the recurrence

$a_{n+1}=\dfrac{a_n}{(n+1)^2}$

We can solve this by recursively substituting the right hand side:

$a_n=\dfrac{a_{n-1}}{n^2}=\dfrac{a_{n-2}}{n^2(n-1)^2}=\cdots=\dfrac{a_0}{(n!)^2}$

and so one of the solutions will be

$y_1(x)=\displaystyle a_0\sum_{n\ge0}\frac{x^n}{(n!)^2}$

Now the literature on the Frobenius method when the indicial polynomial has repeated roots suggests a second solution will take the form

$y_2(x)=Cy_1\ln x+\displaystyle\sum_{n\ge0}b_nx^n$

but it's not clear to me *why* this works.