First Year ODE Exam Question 
What are the eigenfunctions that satisfy
  $$xy''(x) + (1-x)y'(x) + y(x) = -\lambda y(x)$$
  on the interval $I = [0, \infty)$ given that the eigenvalues are natural numbers (positive integers) such that $y(0) = 1$?

This question was on my first-year university ODE final exam and not a single student was able to solve it. We are trying to have it curved from the final exam grade, do we have a strong case or is this fairly "ordinary" material?
 A: I believe, if you have encountered this as a Freshman, you are supposed find the solution as a power series. Quite generally when you have a differential equation of the form
$$
N(x) y'' + M(x)y'+P(x)y=0
$$
with $N,M,P$ polynomials, and $N(x)$ with no roots in the interior of the domain you want to solve your ODE in, then your solution can be written as a power series $y(x)=\sum_{n=0}^\infty a_n x^{n}$ (you don't know at this stage what the coefficient $a_n$ are going to be). I believe this is taught to students rather early, no?

Let's write the ODE slightly differently
$$
\underbrace{xy''}_A+\underbrace{y'}_{B}-\underbrace{xy'}_{C}+\underbrace{m y}_D=0
$$
with $m$ a positive integer (here $m=1+\lambda$). Insert the power series for $y(x)$. You get
$$
\underbrace{\sum_{n=2}^\infty a_n n(n-1)x^{n-1}}_A
+\underbrace{\sum_{n=1}^\infty a_n nx^{n-1}}_{B}-\underbrace{\sum_{n=1}^\infty a_n n x^n}_{C}+\underbrace{m\sum_{n=0}^\infty a_n x^n}_D=0
$$
Or upon manipulation and rearranging becomes
$$
m + a_1+\sum_{n=1}^\infty \Big[(n+1)^2a_{n+1}+(m-n)a_n\Big]x^n=0
$$
The condition $y(0)=1$ gives you $a_0=1$. To find the rest of the coefficients, note that a power series is zero if the coefficient of each power is zero. This means
$
a_1=-m$
and
$$
(n+1)^2a_{n+1}=-(m-n)a_n
$$
Now note that eventually (in fact for $n=m$) the right hand side of the above equation vanishes. This results in $a_{m+1}=a_{m+2}=\cdots = 0$. In other words, your solution is a polynomial. More explictly, you have a a recursive relation (for $n\leq m$)
$$
a_n = -\frac{m -n +1}{n^2}a_{n-1}\Longrightarrow
a_n = (-1)^{n-1}\frac{(m-1)!}{n!^2(m-n)!}a_1=(-1)^n {m\choose n}\frac{1}{n!}
$$
In other words, putting all of these together the solution is
$$
\boxed{
y(x):=L_m(x) = \sum_{n=0}^m (-1)^n {m\choose n}\frac{x^n}{n!}}
$$
This polynomial, as people mentioned in comments is called $m$th Laguerre polynomial, but you are probably not expected to know the name. Finding the above formula should have been enough to get you full credit probably.
