There is a potential $U(x)$ defined by :
$$U(x) = \begin{cases} U_1 & \text{if $x>0$} \\[2ex] U_2, & \text{if $x<0$} \end{cases}$$ such that $U_2>U_1$.
The Schrodinger equation witch rules the system is: $$\frac{d²\psi(x)}{dx²}=(U_n-\epsilon)\psi $$ where $n$ may be $1$ or $2$ depending on the region and $\epsilon>U_2$.
I saw in the book "Quantum Mechanics of Albert Messiah" that the general form of the $\psi$ function in this case is: $$\psi(x) = \begin{cases} \exp(-ik_1x)+R\exp(ik_1x) & \text{if $x>0$} \\[2ex] S\exp(-ik_2x), & \text{if $x<0$} \end{cases}$$ such that $R$ and $S$ are constants to be defined,$k_1²=U_1-\epsilon$ and $k_2²=U_2-\epsilon$.
But by my solution there is more constants to be defined. I arrived in this by the steps bellow: Solving the Schrodinger equation in the two regions I found that the general solution is: $$\psi(x) = \begin{cases} A\exp(-ik_1x)+R\exp(ik_1x) & \text{if $x>0$} \\[2ex] S\exp(-ik_2x)+B\exp(ik_2x), & \text{if $x<0$} \end{cases}$$ Such that $A$ and $B$ are constants to be defined.
I would like to know, why the coefficient $A$ may be 1 and the coefficient $B$ may be 0 like the Messiah affirms.
$A$ and $B$ are both defined from the initial condition and normalization condition. This doesn't come from mathematics, but rather from physics.
Look at this diagram:
We have ourselves decided that there's a particle in a $\mathbf{p}$ state (meaning precisly defined momentum and a wave function in the form of a free wave) coming from the right.
We have set the factor before the exponential to be $1$, because if we compute the probability current, it will be equal to $\mathbf{v}$, the classical velocity of the particle.
Finally, we know that if a travelling wave encounters a change in potential it will partially reflect and partially transmit (or scatter, though the latter term is usually used when the number of dimensions is greater than $1$).
Since there's no wave coming from the left, $B$ should be $0$.