I must find the eigenvalues and eigenfunction for $$y''+2y'=\lambda y$$ with initial conditions $y(0)=0$, $y'(1)=0$. I have found the non-trivial case, and made an attempt to solve for $\lambda$, but I have now been stuck for a long time, and see no way to proceed. Any help would be appreciated.
My attempt:
Let $y=e^{mx}$. Then, we obtain $m^2+2m-\lambda=0 \implies m=\frac{-2 \pm \sqrt{4+4\lambda}}{2}=-1 \pm \sqrt{1+\lambda}$. The non-trivial case is when $1+\lambda <0$, and we have complex roots with $m=-1 \pm \sqrt{-1-\lambda}i$ Then, the general solution will be: $$y(x)=c_1e^{-x}\cos(\sqrt{-1-\lambda}\cdot x)+c_2e^{-x}\sin(\sqrt{-1-\lambda}\cdot x)$$ Plugging in initial conditions, I obtain that $c_1=0$. Now, I take the derivative: $$y'(x)=-c_1e^{-x}\cos(\sqrt{-1-\lambda}\cdot x)-\sqrt{-1-\lambda}\cdot c_1e^{-x}\sin(\sqrt{-1-\lambda}\cdot x)-c_2e^{-x}\sin(\sqrt{-1-\lambda}\cdot x)+\sqrt{-1-\lambda}\cdot c_2e^{-x}\cos(\sqrt{-1-\lambda}\cdot x)$$ Plugging, in initial conditions and $c_1=0$, we have $$c_2\left(\sqrt{-1-\lambda}\cdot e^{-1}\cos(\sqrt{-1-\lambda})-e^{-1}\sin(\sqrt{-1-\lambda})\right)=0$$ Since $c_2=0$ leaves us with a trivial case, the inside of the parentheses must equal zero. Then, cancelling out $e^{-1}$ and rearranging, we have $$\sqrt{-1-\lambda}\cdot \cos(\sqrt{-1-\lambda})=\sin(\sqrt{-1-\lambda})$$ $$\sqrt{-1-\lambda}=\tan(\sqrt{-1-\lambda})$$ No matter what identities I use, or how I rearrange this, I cannot seem to solve for $\lambda$ here. In the past, I have used Euler's Identity, $e^{i \pi}+1=0$ and $e^{i \theta}=\cos(\theta)+i\sin(\theta)$ to help simplify, but they do not seem useful in this case. Any ideas on how to solve this? Thank you.