$Au_{xx} + Bu = 0 $ then $u = 0$ 
Let $A,B > 0$ and $0 < a < b < \infty$. Consider $u \in C^{1}([0,b])$, $u = 0$ in $[0,a)$ and
$$
Au_{xx} + Bu = 0 \ \  \text{in} \ \ (a,b)
$$
with $u(a) = u(b) = 0$. Then $u = 0$ in $[a,b)$.

My ideia: How $u \in C^{1}([a,b])$ and $u= 0$ in $[0,a)$, then $u(0) = u (a) = u_{x}(a) = 0$. By system, we have
$$
u(x) = c_{1}\cos\bigg(\sqrt{\frac{A}{B}}x\bigg) + c_{2}\sin\bigg(\sqrt{\frac{A}{B}}x\bigg), \ \ \text{in} \ \ (a,b)
$$
But, $u(0) = u (a) = u_{x}(a) = 0$, then
$$
u(x) =  c_{2}\sin\bigg(\sqrt{\frac{A}{B}}x\bigg)
$$
with
$$
c_{2}\cos\bigg(\sqrt{\frac{A}{B}}a\bigg) = c_{2}\sin\bigg(\sqrt{\frac{A}{B}}a\bigg) = 0
$$
Soon, $\frac{A}{B}a = \frac{\pi}{4} + \pi n$ or $c_{2} = 0$. And now? I can't continue.
 A: Since the solution $u(x) = c_1 \cos (\sqrt{\frac{B}{A}} x) + c_2 \sin (\sqrt{\frac{B}{A}} x)$ holds for $x \in (a,b)$, the fact that $u(0)=0$ is not a constraint on the coefficients but rather a data given by the problem $u(x) = 0$ for $x \in [0,a)$
All you can impose is
$$ u(a) = u_x(a) = u(b) = 0$$
By translation symmetry, it is then more convenient to use the solution
$$ u(x) = c_1 \cos \left(\sqrt{\frac{B}{A}} (x-a)\right) + c_2 \sin \left(\sqrt{\frac{B}{A}} (x-a)\right)$$ (which is equivalent to your solution by some reparametrization of the coefficients)
The conditions at $x=a$ now imply
$$ u(a) = c_1= 0 $$
$$ u_x(a) = \sqrt{\frac{B}{A}} c_2 = 0 $$
Then we see that both coefficients vanish $c_1 = c_2 = 0$ hence $u=0$.
Notice that we didn't have to use what happens at $b$. The only knowledge that both the function and its derivative are 0 at some point tells you that it can not be a sum of sines and cosines.

NB: If we really wanted to use the solution $u(x) = c_1 \cos (\sqrt{\frac{B}{A}} x) + c_2 \sin (\sqrt{\frac{B}{A}} x)$, the condition would be
$$u(a) = c_1 \cos \left(\sqrt{\frac{B}{A}} a\right) + c_2 \sin \left(\sqrt{\frac{B}{A}} a\right)=0$$
$$u_x(a) = \sqrt{\frac{B}{A}} \left(-c_1 \sin \left(\sqrt{\frac{B}{A}} a\right) + c_2\cos \left (\sqrt{\frac{B}{A}} a \right) \right) = 0$$
Removing the square root prefactor, and summing the squares of these two equations, we would get
$$ c_1^2 + c_2^2 = 0$$
which holds only when both coefficients vanish
