Help on an eigenvalue problem?

[Robam]

Hello, I'm stuck on a fairly straightforward eigenvalue problem. For
y'' + 2y' + λy = 0; y(0) = y(1) = 0
(a) Show that λ = 1 is not an eigenvalue.
(b) Show that there is no eigenvalue λ such that λ < 1
(c) Show that the nth positive eigenvalue is λ_n = n²π² + 1

I've already shown (a) and (b). On (c), I used the characteristic equation and assumed λ > 1 to find that
y = e^-x (c₁ cos((1-λ)^1/2 x) + c₂ sin((1-λ)^1/2 x))
From there, I found c₁ = 0, and e^-1 c₂ sin((1-λ)^1/2) = 0
For a nontrivial solution, I set sin((1-λ)^1/2) = 0
So λ = -n²π² + 1

Why am I off by a negative? There's probably something simple I'm missing, but I can't see it.

 

[HallsofIvy]

With \(\displaystyle \lambda> 1\), \(\displaystyle 1- \lambda< 0\) so that \(\displaystyle \sqrt{1- \lambda}\) is imaginary.
Since you want real number coefficients, you need to write that as \(\displaystyle \sqrt{1- \lambda}= \sqrt{-1(\lambda- 1)}= i\sqrt{\lambda- 1}\).


So what you really have, as \(\displaystyle \theta\) in \(\displaystyle e^{i\theta}= cos(\theta)+ i sin(\theta)\), is \(\displaystyle \theta= \sqrt{\lambda- 1}\) so you should have \(\displaystyle \sqrt{\lambda- 1}= n\pi\), \(\displaystyle \lambda- 1= n^2\pi^2\), \(\displaystyle \lambda= n^2\pi^2+ 1\)