# Finding the solution to a DE with complex roots

I can't figure out how to get the answer in the back of the book.

y''+6y'+13y=0

So far, I have.....

ert (r2 + 6r + 13) = 0
r2 + 6r + 13 = 0
r2 + 6r + ___ = -13 + ___
r2 + 6r + 9 = -13 + 9
(r+3)2 = -4
r = -3 +/- 2i

Then we have a crazy-looking thing after assuming y = C1e(r1)(t) + C2e(r2)(t)

y = e-3t (C1e(2t)i + C2e(-2t)i

And using Euler's formula and simplifying a little bit

y = e-3t ( C1 ( cos(2t) + i*sin(2t) ) + C2( cos(-2t) i*sin(-2t) ) )

The answer in the back of the book, however, is something that doesn't even involve imaginary numbers.

What am I missing?

Pengwuino
Gold Member
What was the answer in the back of the book that you are finding conflict with? Looks right to me assuming you meant the $$e^{-3t}$$ to multiply the whole thing in yoru last line.

What was the answer in the back of the book that you are finding conflict with? Looks right to me assuming you meant the $$e^{-3t}$$ to multiply the whole thing in yoru last line.

y=C1e-3tcos(2t) + C2e-3tsin(2t)

There's probably some trig identity eluding me I'm pretty sure the answer from the book is just a rearranged form of your equation
y = e-3t ( C1 ( cos(2t) + i*sin(2t) ) + C2( cos(2t) - i*sin(2t) ) )

Just multiply out everything, then you can factor the sine and cosine terms, so it's of the form e-3tcos(2t)+e-3tsin(2t). The rest is just renaming constants, the book solution just absorbed the i into their C2.