Consider the second-order differential equation
If is a point in the plane and if
is a given real number, prove that this differential equation has exactly one solution whose graph passes through the point
and has slope
. Consider also the case
.
Proof. If , then
implies
, and
implies
. This gives us
. Then,
so
If , then
so
. This means the solutions are of the form
The condition then gives us
The condition give us
Thus, we have a unique solution with
and
as given above
I think you have to consider the case ka=(2n+1)pi/2
Imo too, since if that is the case, then the division by cos(ka) is an invalid step.
Also when substituting
into
the sign of
should be – not +.