Uniqueness and Existence of $u^{\prime\prime}(x) + u^{\prime}(x) = f (x)$ with conditions. Here is my question:
Given
$u^{\prime\prime}(x) + u^{\prime}(x) = f (x)$ with the conditions
$u^\prime(0) = u(0) = {1\over 2}[u^\prime(l) + u(l)]$
Where $f(x)$ is a given function, is the solution unique? Does a solution necessarily exist, or is there a condition that f (x) must satisfy for existence?

The back of our text provides suggestions for some problems. It suggested for the first part (uniqueness) to find two solutions, and take the difference, then solve the ODE. For the second part, it suggested to integrate the equation from $0$ to $l$.
For the first part, I seem to be having an issue finding more than one solution for a given $f(x)$. I have been working with $f(x)=0$, and therefore having the trivial solution of $u(x)=0$. Other than that though, I have been unable to find a solution to work with that works with the conditions (closest I found was $u(x)=-e^{-x}$.
Any help would be appreciated - thanks.
 A: Update: There was an error in the calculation.
We first solve the IVP
$$u''+u'=f(x),\qquad u(0)=u'(0)=0\ .$$
The auxiliary function $v(x):=e^x u'(x)$ satisfies
$$v'(x)=e^x f(x),\qquad v(0)=0\ .$$
It follows that
$$v(x)=\int_0^x e^\tau f(\tau)\>d\tau$$
and then
$$u(x)=\int_0^x e^{-t} v(t)\>dt=\int_0^x \left( e^{-t} \int_0^t e^\tau f(\tau)\>d\tau\right)\>dt$$
This particular solution $u_0$ can after some manipulation of integrals be rewritten as
$$u_0(x)=\int_0^x\bigl(1-e^{\tau-x}\>f(\tau)\bigr)\>d\tau\ .$$
Therefore we obtain the following general solution of the original ODE:
$$u(x)=c_0+ c_1e^{-x}+u_0(x)\qquad(0\leq x\leq \ell)\ .$$
Now we have to take care of the boundary conditions
$$c_0+c_1=-c_1={1\over2}\bigl(c_0 +C\bigr)\ ,\tag{1}$$
where
$$C=u_0'(\ell)+u_0(\ell)=e^{-\ell}v(\ell)+u_0(\ell)=\int_0^\ell f(\tau)\>d\tau\ .$$
The first condition $(1)$ enforces $c_1=-{c_0\over 2}$, and then the second condition can only be satisfied if $C=0$, whereby $c_0$ may be arbitrary. Therefore we have the following alternative:
If $C:=\int_0^\ell f(\tau)\>d\tau\ne0$ the given boundary value problem has no solution. If $C=0$ the problem has infinitely many solutions given by
$$u(x)=c_0\left(1-{1\over2}e^{-x}\right)+u_0(x),\qquad c_0\in{\mathbb R}\ .$$
A: Using the suggestion, let $u_1, u_2$ be two solutions. Then $u_1-u_2$ satisfies
$$u''(x)+u'(x)=0$$
By solving the characteristic equation $r^2+r=0$, we get $u_1(x)-u_2(x)=c_1 e^{-x}+c_2$.
The boundary condition $u^\prime(0) = u(0) = {1\over 2}[u^\prime(l) + u(l)]$ gives us 
$$u_1^\prime(0)-u'_2(0) = u_1(0)-u_2(0) = {1\over 2}[(u_1-u_2)^\prime(l) + (u_1-u_2)(l)]$$
Plugging these into our $u_1-u_2$, we get $c_1=-\frac{1}{2}c_2$ for both. 
So $u_1=u_2+C(1-\frac{1}{2}e^{-x})$. You can check that $1-\frac{1}{2}e^{-x}$ is not a solution. So $u_1, u_2$ are two different solutions.
