The first three stationary states of a particle in the potential V= g(1-(2/(1+((Tanh[λ x])^2))))

The code below solves the time-independent Schroedinger equation for the potential  V=10(1-(2/(1+((Tanh[λ x])^2))))  with the boundary condition that ψ[-∞]=0  (and an arbitrary normalization) and an energy En. For a random value of En the solution will not be zero at x=∞ and is not  acceptable. For certain magic values of En the solution will satisfy  ψ[∞]=0  and is acceptable. Those are the energy eigenvalues of the system.

En=;(* put your guess for En here *)
even=NDSolve[{-((hbar^2)/(2 m))ψ''[x]+V[x] ψ[x]==En ψ[x], ψ[-inf]==0, ψ'[0]==5}, ψ, {x, -inf, inf}]; (* this finds even solutions *)
Plot[{Evaluate[ψ[x]/.even], V[x]},{x,-inf, inf},PlotRange->All]

hbar=1; (* we use a funny unit system here where hbar=1 *)
m=1;
inf=6; (* the width of the potential is about ~1 and "infinity" is taken, for numerical reasons, to be 6. Any other number much larger than the width of the potential will do. *)
λ=1. ;V[x_]:=10(1-(2/(1+((Tanh[λ x])^2))));
En=;(* put your guess for En here *)
even=NDSolve[{-ψ''[x]+((2m)/(hbar^2))V[x] ψ[x]==((2m)/(hbar^2))En ψ[x], ψ[-inf]==0, ψ[0]==5.}, ψ, {x, -inf, inf}]; (* this finds even solutions *)
Plot[{Evaluate[ψ[x]/.even], V[x]},{x,-inf, inf},PlotRange->All]

En=;(* put your guess for En here *)
even=NDSolve[{-ψ''[x]+((2m)/(hbar^2))V[x] ψ[x]==((2m)/(hbar^2))En ψ[x], ψ[-inf]==0, ψ'[0]==10.}, ψ, {x, -inf, inf}]; (* this finds odd solutions *)
Plot[{Evaluate[ψ[x]/.even], V[x]},{x,-inf, inf},PlotRange->All]