Why Do Solutions to y'' + a²y = 0 Include Complex Conjugates?

[Unto]

The general solution of the second order DE

y'' + a^2y = 0

is

y = A cos ax + B sin ax

WTF?



Someone explain why?

 

[Mark44]

The characteristic equation is r² + a² = 0, which has solutions r = +/- ai. This means that the general solution is all linear combinations of {e^iax, e^-iax} = {cos ax + i sin ax, cos ax - i sin ax). By taking suitable linear combinations of these two complex functions, you can get two real solutions, sin(ax) and cos(ax).

 

[Unto]

Why does the linear combinations include complex conjugate?

 

[Mark44]

A 1st order differential equation has 1 basic solution, a 2nd order differential equation has 2 basic solutions, a 3rd order differential equation has 3 basic solutions, and so on.

The basic solutions are all of the form e^rx, where r is a root of the characteristic equation. If the characteristic equation has complex roots, these roots always come in pairs - the complex conjugates. For the sake of convenience, instead of writing e^{(a + bi)x} and e^{(a - bi)x}, we do a little algebra and write these functions as e^axcos(bx) and e^axsin(bx) so that we don't have to mess with imaginary numbers at all.

 

[Unto]

Yes I understand that, I am proficient with complex numbers.

 

[Mark44]

So did I answer your question?

 

[crd]

Why does the linear combinations include complex conjugate?

because the solutions form a 2-dim vector space over the field of complex numbers, so by properties of vector spaces all scalar multiples are elements of said solution space, which in this case would include linear combinations of complex conjugates