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.

Because the expression for solving for m is a quadratic eq. and will always have two solutions which are complex conjugates of each other. Because of this, all solutions for second order linear eq. like this (with compex roots and with a non-zero real part) will have some exponential multiplying Ae^(c*ix)+Be^(-c*ix) (in your case c=1).

Either I missed the point of your answer, or you missed the point of my question.

I know that it has some exponential multiplied by that expression, but my question relates to the lines below it. There is an unexplained step in getting from A+B and i(A-B) to a and b respectively, justified by A and B being conjugate pairs. But what I don't understand is why or how they are in fact conjugate pairs.

Ah, I missed the point of your question. From Euler's formula:

Sin(x) = ( e^(ix)-e(-ix) )/(2i)
Cos(x) = ( e^(ix)+e^(-ix) )/2

a*sin(x) + b*cos(x) =

a*( e^(ix)-e(-ix) )/(2i) + b*( e^(ix)+e^(-ix) )/2=

(Factoring)
=[b/2 + a/(2i)]*e^(ix) +[b/2 - a/(2i)]*e^(-ix)

=Ae^(ix)+Be^(-ix),
where A=b/2 - i*a/2 and B=b/2 + i*a/2

That is the algebra that links a and b to A and B, and shows that A and B are complex conjugates. Since they are all arbitrary variables, any time A and B are complex conjugates, you may assume it is in the right form and find the values for a and b.

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.

>>5
Speaks tr00f. We assume that the solution plane of such a differential equation is real, since we model physical processes with differential equations, and it's a little hard for a wave to be (2 + 3i)m high.

nerd alert

jk, you guys sound rather intelligent

MATH IS HARD

The word pirahna, is all I can think of that rhymes with marijuana

Marijuana MUST be legalized.

The word pirahna, is all I can think of that rhymes with marijuana

Marijuana MUST be legalized.