Finding the eigenvalues and eigenvectors I'm trying to determine if I did this problem properly.
Give the following ODE $y''-2 y'+(\lambda-1)y=0$ with $y(0)=0=y(1)$.
Find the eigenvalues and eigen-functions.
To do this, I rewrote
$$r^2-2r+1=1-\lambda+1 \Rightarrow r=1 \pm \sqrt{2-\lambda}.$$ 
I reduced this to three cases: $\lambda <2,\lambda =2, \lambda>2$.
Case 1:  $\lambda<2$. 
In this case we have 
$$y=C_1e^{(1+\sqrt{2-\lambda})x}+C_2e^{(1-\sqrt{2-\lambda})x}$$. 
Using $y(0)=0$ the initial conditions I have $C_1=-C_2$.  Using the second initial condition of $y(1)=0$, we now have 
$$0=-C_2e^{(1+\sqrt{2-\lambda})}+C_2e^{(1-\sqrt{2-\lambda})}$$.
Some algebra later, I get $\lambda =2$.
Case 2: $\lambda=2$
In this case we have $y=C_1e^x+C_2xe^x$.  From this, we get the trivial solution.
Case 3: $\lambda > 2$
In this case I have 
$$y=C_1e^x\cos((\sqrt{2-\lambda}x))+C_2e^x\sin((\sqrt{2-\lambda}x)).$$
With $0=y(0)=C_1.$  Also, $0=y(1)=C_2e\sin((\sqrt{2-\lambda})).$  
Moving things around we have 
$$\arcsin(0)=\sqrt{2-\lambda} \Rightarrow \pi k=\sqrt{2-\lambda}=(\pi k)^2=2-\lambda\Rightarrow 2-(\pi k)^2=\lambda$$.  
This leaves us $\lambda_k=c_k\sin(\pi k)$.
So our eigenvalues are $\lambda=2-(\pi k)^2$ and our eigen-functions are:
$$y_k=c_k\sin(\pi k)$$
where $k=0,1,\dots$.
 A: One way to save a little effort by eliminating special cases is to solve for the unique $y$ such that $y(0)=0,y'(0)=1$, and then plug in at $x=1$ to find the eigenvalue equation. In this case,
$$
       y(x)=C_1e^{(1+\sqrt{2-\lambda})x}+C_2e^{(1-\sqrt{2-\lambda})x}
$$
may be written as
$$
            y(x)=D_1 e^{x}\sin(\sqrt{\lambda-2}x)+D_2e^{x}\cos(\sqrt{\lambda-2}x),
$$
which makes it easy to solve $y(0)=0$, $y'(0)=1$. Automatically $D_2=0$, and $y'(0)=1$ gives
$$
           y(x)=e^{x}\frac{\sin(\sqrt{\lambda-2}x)}{\sqrt{\lambda-2}}.
$$
This eliminates special cases because taking the limit as $\lambda\rightarrow 2$ also gives the correct solution for $\lambda=2$ as $y(x)=xe^x$. Now the eigenvalue equation becomes
$$
    y(1)=0\implies \frac{\sin(\sqrt{\lambda-2})}{\sqrt{\lambda-2}}=0 \tag{$\dagger$}
$$
When you do it this way, you always end up with a power or McClaurin series in $\lambda$ for the eigenvalue equation, which eliminates special cases such as $\lambda=2$. For example, in this case,
$$
          0 = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}(\lambda-2)^{n}.
$$
But you don't have to write out the series. Just know that there is one, and the special cases are limiting cases. You can see from $(\dagger)$ that the limiting case of $\lambda=2$ does not work, so that one is eliminated. The special case at $\lambda=2$ is no longer a special case. The other solutions are $\sqrt{\lambda-2}=n\pi$ for $n\ne 0$.
