Uniqueness of solution for second order differential equation with two Dirichlet conditions Given is the differential equation $-cu''(x) + (u'(x))^2 - 1 = 0$ in $(-1,1)$ and $u(-1)=u(1)=0$ for some constant $c>0$.
I need to show that there exists a unique solution $u\in C^2(-1,1)\cap C([-1,1])$ for the Dirichlet problem given above. I already know that $u(x)= -c \log\left(\frac{ e^{\frac{x}{c}} + e^{-\frac{x}{c}}}{e^{\frac{1}{c}} + e^{-\frac{1}{c}}} \right)$ is a solution, i just can't show that it is unique. I have also tried it via contradiction but it didn't work. Has anybody got an idea in which book or paper to differential equations one might find a theorem on the uniqueness of a solution of second order differential equations? I only found some theorems for problems with one Dirichlet condition and one condition for the first derivative but for two Dirichlet conditions and no condition for the first derivative I just could not find anything!
Thank you all for your help, best regards!

Assume $u,v:[-1,1] \to \mathbb{R}$ are different solutions of this BVP. W.l.o.g. assume that $v(\xi)< u(\xi)$ for some $\xi \in (-1,1)$. Choose
$\varepsilon > 0$ minimal such that
$$
v_\varepsilon(x):=v(x)+ \varepsilon \ge u(x) \quad (x \in [-1,1]).
$$
Since $\varepsilon$ is minimal there is some $x_0 \in (-1,1)$ such that
$v_\varepsilon(x_0)=u(x_0)$. Since $v_\varepsilon - u \ge  0$ on $[-1,1]$ and $v_\varepsilon(x_0) - u(x_0)= 0$ we have $v'_\varepsilon(x_0)=u'(x_0)$.
Since $v_\varepsilon$ is a solution of the differential equation both functions $u, v_\varepsilon$ are solutions of the same initial value problem at $x_0$. Hence $v_\varepsilon = u$, a contradiction (as $v_\varepsilon(1) \not= u(1)$).
