What are the eigenvalues of this problem?

Question

I have the following problem that has been giving me a lot of trouble:

$$y''-3y'+3(1+\lambda)y=0$$ $$y'(0)=y'(1)=0$$

I know how to solve this type of eigenvalue problems. I try $y=e^{rx}$ and the characteristic polynomial gives me: $r^2 -3r+3(1+\lambda)=0$, so that $$r=\frac{3\pm\sqrt{9-12(1+\lambda)}}2$$, so that the cases I have to study are:

1. $\lambda=\frac{-1}4$
2. $\lambda<\frac{-1}4$
3. $\lambda>\frac{-1}4$

I go on to solve this and obtain my eigenvalues, but my answer is completely different from the book's numerical solution, in which they have found the eigenvalue $\lambda_0=-1$. My question is, how can this be? How do they get to this case? I just need to know this, because I have spent a lot of time on this problem and it is getting me nowhere.

Edit: I have figured out what I was missing in my solution. Thanks for all your help!

Answer 1

Everything is correct - the general form of the differential equation is $y=e^{rx}$, and when $\lambda\le-1/4$, you get the solution for r.

Now, you still have to solve

$y'(0)=y'(1)=0$

and this will give you more information about r / about $\lambda$.

Hope this helps!