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

It’s not needed with some trickery. b-bsin^2(ka) = bcos^2(ka) so we can cancel the cosine from the denominator. Also k should be in absolute value since nobody said it’s positive. If you don’t like this method, when solving for the constants, multiply both sides by sine or cosine to get sin^2 in one equation and cos^2 in the other and when you add them the coefficient becomes 1.

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 +.

It’s not needed with some trickery. b-bsin^2(ka) = bcos^2(ka) so we can cancel the cosine from the denominator. Also k should be in absolute value since nobody said it’s positive

It’s not needed with some trickery. b-bsin^2(ka) = bcos^2(ka) so we can cancel the cosine from the denominator. Also k should be in absolute value since nobody said it’s positive. If you don’t like this method, when solving for the constants, multiply both sides by sine or cosine to get sin^2 in one equation and cos^2 in the other and when you add them the coefficient becomes 1.