Say we have the following equation the following equation (inspired from quantum mechanics with $V\left(x\right) = - e^{- \alpha x}$):
\begin{align} \label{eq:first-eq} - \frac{\partial^2 \psi\left(x\right)}{\partial x^2} - e^{-\alpha x} \psi =0 \tag{1} \end{align}
According to Mathematica the solutions are linear combinations of Bessel functions:
\begin{align} \label{eq:second-eq} \psi\left(x\right) = A J_{0} \left(\frac{2}{\alpha }e^{-\alpha x/2} \right) + B Y_{0} \left(\frac{2}{\alpha }e^{-\alpha x/2} \right) \tag{2} \end{align}
The code:
eq = -D[ψ[x], {x, 2}] - E^(-α*x)*ψ[x] == 0
DSolve[eq, ψ[x], x]
{{ψ[x] ->
BesselJ[0, (2 Sqrt[E^(-x α)])/α] C[1] + 2 BesselY[0, (2 Sqrt[E^(-x α)])/α] C[2]}}
However, some is strange about this solution. Specifically notice that if we consider the equation:
$$- \frac{\partial^2 \psi\left(x\right)}{\partial x^2} - e^{\alpha x} \psi =0 $$
Which is Eq. \ref{eq:first-eq} but with $\alpha \rightarrow -\alpha $, then we should get Eq. \ref{eq:second-eq} with the $\alpha \rightarrow -\alpha $. However, instead we get:
$$ \psi\left(x\right) = A J_{0} \left(\frac{2}{\alpha }e^{\alpha x/2} \right) + B Y_{0} \left(\frac{2}{\alpha }e^{\alpha x/2} \right) $$
eq = -D[ψ[x], {x, 2}] - E^(α*x)*ψ[x] == 0
DSolve[eq, ψ[x], x]
{{ψ[x] ->
{{ψ[x] -> BesselJ[0, (2 Sqrt[E^(x α)])/α] C[1] + 2 BesselY[0, (2 Sqrt[E^(x α)])/α] C[2]}}
I know that $J_{0}$ is even but $Y_{0}$ is not, what is happening? Also, which of the solutions is actually correct?