The solution in region II must solve the differential equation (2) and match the appropriate boundary conditions. The boundary conditions at x=-a, as discussed in the previous section, are that the slope and value of the wavefunction at x=-a must match those of the wavefunction in region I,
In region I the TISE reads
where we have introduced a new constant
with a minus sign in (10) to remind us of the sign of the left hand side of (2).
Just as in region I, this is a second order linear differential
equation with constant coefficients, and so again, this equation must
also have solutions of the form 
. In this case
. The general solution to (10) is
thus 
, where 
and 
are the two independent constant parameters. Again, it is
good form to write the solution as centered at the boundary
where 
and 
. In this case, one final redefinition
of constants is convenient. Because the TISE is a real differential
equation, we expect it to possesses two linearly independent real
solutions. We bring this out explicitly by writing new constants
and 
giving us the form
Again, one may arrive at 12 directly. In a
classically allowed region
(where we define k's instead of 
's), the solutions are
always sinusoidal and direct substitution verifies that
(12) solves (10).
With the most general solution to the differential equation region II, it remains only to adjust the coefficients so that the boundary conditions are matched. With the function written in this form, the algebra of this is simple,
and
Comparing this with with the value and
slope carried across, ``injected'' from region I (9) we
have 
, 
. With the constants
determined, we have the final solution in region II,
