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 +.