Finding the solution to a DE with complex roots

[Jamin2112]

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

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

So far, I have...

e^rt (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]

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.

 

[Jamin2112]

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

 

[Brian-san]

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.