Solving $u''-4u=\lambda u$ with Sole Boundary Condition $u'(0)=u(\pi)$ 
Find the eigenfunctions and eigenvalues of the boundary value problem,  $$u''-4u=\lambda u, \ \ \ 0<x<\pi$$ with $u'(0)=u(\pi)$. Check that the resulting eigenfunctions are orthogonal.

I have solved many problems similar to this, but with more than one boundary condition. I don't see how this problem is solved given this single condition.
My attempt:
\begin{align}
u''-u(4+\lambda)&=0 \\
u''-u\mu&=0, \ \ \text{where $\mu=4+\lambda$.} \\
\end{align}
Consider values of $\mu$. If $\mu=0$, then 
\begin{align}
u''&=0 \\
u(x)&=Ax+B \ \ \  \ A,B\in\mathbb{R} \\
u(x)&=Ax+A(1-\pi) \ \ \ \text{(using boundary condition)}.
\end{align}
Continuing in this way, I found no trivial solutions for values of $\mu$? How can this be?
 A: Clearly if $\mu=0$, the eigenproblem is trivial. For $\mu\neq0$, the corresponding characteristic equation is given
$$ r^2-\mu =0$$
which has two roots $r=\pm\sqrt{\mu}$ if $\mu>0$ or $r=\pm\sqrt{-\mu}i$ if $\mu<0$. Then the general solution of
$$ u''-u\mu=0 $$
is 
$$u=Ae^{\sqrt{\mu}x}+Be^{-\sqrt{\mu}x} $$
if $\mu>0$ or
$$u=A\sin(\sqrt{-\mu}x)+B\cos(\sqrt{-\mu}x) $$
if $\mu<0$. 
Case 1: $\mu>0$. Using $u'(0)=u(\pi)$, one has 
$$ \sqrt{\mu}A-\sqrt{\mu}B=0, Ae^{\sqrt{\mu}\pi}+Be^{-\sqrt{\mu}\pi}=0$$
From this, it is easy to see that $A=B=0$; namely, the problem does not have nonzero eigenfunction.
Case 1: $\mu>0$. Using $u'(0)=u(\pi)$, one has 
$$ \sqrt{\mu}A-\sqrt{\mu}B=0, Ae^{\sqrt{\mu}\pi}+Be^{-\sqrt{\mu}\pi}=0$$
From this, it is easy to see that $A=B=0$; namely, the problem does not have nonzero eigenfunction.
Case 2: $\mu<0$. Using $u'(0)=u(\pi)$, one has 
$$ A\sqrt{-\mu}=0, A\sin(\sqrt{-\mu}\pi)+B\cos(\sqrt{-\mu}\pi)=0$$
From this, it is easy to see that $A=0$. In order to seek nonzero eigenfunction, set
$$ \sqrt{-\mu}=n $$
where $n=1,2,3,\cdots$. So $\mu=-n^2$ or $\lambda=-n^2-4$ and the corresponding eigenfunction is $u_n=B\cos(n x)$. It is easy to check that
$$ \int_0^\pi\cos(mx)\cos(nx)dx=0, m\neq n$$
and
$$ \int_0^\pi\cos^2(mx)dx=\frac\pi2. $$
So the eigenfunctions are orthogonal.
