Answer
See the detailed answer below.
Work Step by Step
To prove that the superposition \( \psi(x) = A\psi_1(x) + B\psi_2(x) \) is a solution to the Schrödinger equation, we need use the time-independent Schrödinger equation of
$$
\frac{d^2 \psi_{(x)}}{dx^2} = -\frac{2m}{\hbar^2} [E - U(x)] \psi_{(x)}\tag 1
$$
We know that $ \psi_1(x) $ and $ \psi_2(x) $ are solutions to the time-independent Schrödinger equation for the same potential energy $ U(x)$.
So the superposition is
$$ \psi(x) = A\psi_1(x) + B\psi_2(x)\tag 2 $$
where $ A $ and $ B $ are constants.
The second derivative of $\psi_{(x)}$ with respect to $x$ is given by
$$
\frac{d^2 \psi_{(x)}}{dx^2} = \frac{d^2}{dx^2} \left[A \psi_1(x) + B \psi_2(x)\right]$$
Substituted from (2),
$$
\frac{d^2 \psi_{(x)}}{dx^2} = A \frac{d^2 \psi_1(x)}{dx^2} + B \frac{d^2 \psi_2(x)}{dx^2}
\tag 3 $$
Plugging from (1),
The second derivatives of $\psi_1(x)$ and $\psi_2(x)$ are given by
$$ \frac{d^2 \psi_1(x)}{dx^2} = -\frac{2m}{\hbar^2} [E - U(x)] \psi_1(x) \tag 4$$
$$ \frac{d^2 \psi_2(x)}{dx^2} = -\frac{2m}{\hbar^2} [E - U(x)] \psi_2(x) \tag 5$$
Substituting these expressions from (4) and (5) into (3).
$$
\frac{d^2 \psi_{(x)}}{dx^2} = A \left(-\frac{2m}{\hbar^2} [E - U(x)] \psi_1(x)\right) + B \left(-\frac{2m}{\hbar^2} [E - U(x)] \psi_2(x)\right)
$$
Factoring out $-\dfrac{2m}{\hbar^2} [E - U(x)]$ from both terms:
$$
\frac{d^2 \psi_{(x)}}{dx^2} = -\frac{2m}{\hbar^2} [E - U(x)] \left(A \psi_1(x) + B \psi_2(x)\right)
$$
Noting that $ A \psi_1(x) + B \psi_2(x)=\psi_{(x)} $, from (1), So
$$
\boxed{\frac{d^2 \psi_{(x)}}{dx^2} = -\frac{2m}{\hbar^2} [E - U(x)] \psi_{(x)}}$$
This final expression is identical to the original Schrödinger equation for $\psi_{(x)}$. Therefore, $\psi_{(x)}$, as defined by the linear combination of $\psi_1(x)$ and $\psi_2(x)$, is also a solution to the Schrödinger equation.
