Trigonometric Coefficients of 2nd-order DEs

y'' + 4y' + 5y = 0

as usual, assume y = e^(mx) is a solution, yielding

m^2 + 4m + 5 = 0
m = -2 + i, or m = -2 - i

therefore

y = Ae^((-2+i)x) + Be^((-2-i)x)
= e^(-2x) (Ae^(ix) + Be^(-ix))

by Euler's formula,

y = e^(-2x) ((A+B)cosx + i(A-B)sinx)

as it turns out, A and B work out to be conjugate pairs, thus yielding

y = e^(-2x) (acosx + bsinx)



MY QUESTION: Can anyone explain why A and B are conjugate pairs?

tl;dr - you need to learn more math.

There's a hidden assumption that y is real for all x.

y = Ae^((-2+i)x) + Be^((-2-i)x)
  = e^(-2x)[Ae^ix + Be^-ix]

Assumption only holds if A, B are complex conjugates (if you don't belive me, try stuffing in non complex conjugates, and mucking with x).

If you still don't believe me,

Ae^ix + Be^-ix = (A+B)cosx + i(A-B)sinx (true for any complex A, B)

cos(x), sin(x) are real for all real x, so (A+B) and i(A-B) must be real for the whole thing to be real for all x.  This then implies that A, B are complex conjugates.