Showing that the solution of the in-homogeneous ODE: $u''(x)-Cu(x)=-f(x)$ is unique Given the inhomogeneous ODE on the form: $u''(x)-Cu(x)=-f(x)$
where $0 < x < 1, C > 0$ and $f(x)$ is cont. on the interval for x.
With IC: $u(0) = u(1) = 0$
Solving that yields the general solution for the homogeneous part:
$u_c = Ae^\sqrt{C} + Be^{-\sqrt{C}}$
Inserting initial conditions yields the equality $\sqrt{C} = -\sqrt{C}$ (which can't be since C is strictly positive, right?) since we find that $A = -B$ 
Also how would one go on finding a particular solution when the form of f(x) is unknown? 
Update
If discretizing this system would that be a valid method to prove this has a unique solution, or would that only be valid for the discrete system?
Analytically i guess we would use Variation of parameters as follows:
A solution would be of the form:
$y = y_c + y_p$
An attempt at an solution using variation of parameters yield:
$y_p = y_1u_1+y_2u_2$
$y_1 = e^{(\sqrt{C})x}$
$y_2 = e^{-(\sqrt{C})x}$
$f = f(x)$
$u_1 = \int \dfrac{-y_2 f}{W(y_1,y_2)} dx = -\int \dfrac{e^{-(\sqrt{C})x}f}{1} dx = -\int {e^{-(\sqrt{C})x}f} dx $
$u_2 = \int \dfrac{y_1 f}{W(y_1,y_2)} dx = \int \dfrac{e^{(\sqrt{C})x}f}{1} dx = \int {e^{(\sqrt{C})x}f} dx $
Where (W = Wronskian determinant):
$W(y_1,y_2) = e^{(\sqrt{C}-\sqrt{C})x} = 1$
A solution would be of the form:
$y = y_c + y_p$
$y_c = 0$ so:
$y = y_p = y_1u_1+y_2u_2 = -e^{(\sqrt{C})x} \int {e^{-(\sqrt{C})x}f} dx + e^{-(\sqrt{C})x} \int {e^{(\sqrt{C})x}f} dx $
Inserting for IC yields (all the expontentials cancel out):
$y(0) = y_p = -\int {f(0)} dx + \int {f(0)} dx = 0$
$y(1) = y_p = - \int {f(1)} dx +  \int {f(1)} dx = 0$
So $y = 0$ and does this tell us that the solution is unique? But is 0 a valid solution? This solution differs from the Green kernel method in the answer below, so this solution appears to be wrong.
Now this seems problematic, now am I using the wrong technique or using it wrongly? 

It should of course be
$$
u_c(x) = Ae^{\sqrt{C}x} + Be^{-\sqrt{C}x}
$$
giving you the conditions 
$$
A+B=0\\
Ae^{\sqrt{C}} + Be^{-\sqrt{C}}=0
$$
which indeed tells you that there are no solution except the trivial solution $u_c\equiv 0$.

For the inhomogeneous equation you could either use variation of constants or use the Green kernel to express the same. For the latter, find the solutions with $u(0)=0=v(1)$ and $u'(0)=1=v'(1)$ which are
$$
u(x)=\frac1{\sqrt C}\sinh(\sqrt Cx),\qquad 
v(x)=\frac1{\sqrt C}\sinh(\sqrt C(x-1))
$$
resulting in a Green function
$$
G(x,y)=-\frac1C·\sinh(\sqrt C\min(x,y))·\sinh(\sqrt C(1-\max(x,y)))
$$
for the particular solution
$$
u_p(x)=\int_0^1 G(x,y)f(y)\,dy
$$
