Eigenfunction differential equation with boundary values

Question

Consider the differential equation $$f'' + 2f' + (\lambda + 1)f = 0, \ \ \ \ f(0) + f'(0) = 0, f(L) = 0.$$ We can make $g(x) = e^x f(x)$, so our differential equation becomes $$g'' + \lambda g = 0.$$ Equations like these have nice closed-form solutions. However, the boundary conditions are confusing me. Now we have $$g(0) + g'(0) = f(0), \ \ \ \ g(L) = 0.$$ Since we don't know what $f(0)$, I don't understand how we can possibly solve this. Does anyone know how to solve this?

Answer 1

Since $f(x)=e^{-x}g(x),$ then we have $f'(x)=-e^{-x}g(x)+e^{-x}g'(x)$ and the boundary conditions become

$$f(0)+f'(0)=g(0)-g(0)+g'(0)=g'(0)=0\implies g'(0)=0$$ $$f(L)=e^{-L}g(L)=0\implies g(L)=0$$

Now consider the cases $\lambda <0$, $\lambda=0$ and $\lambda>0$

Answer 2

$$f(0)+f'(0)=e^0f(0)+e^0f'(0)=(e^xf(x))'|_{x=0}$$ $$=g'(x)|_{x=0}=g'(0)=0$$