- #1
Jamin2112
- 986
- 12
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?
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?