• Support PF! Buy your school textbooks, materials and every day products Here!

Finding the solution to a DE with complex roots

  • Thread starter Jamin2112
  • Start date
  • #1
986
9
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?
 

Answers and Replies

  • #2
Pengwuino
Gold Member
4,989
15
What was the answer in the back of the book that you are finding conflict with? Looks right to me assuming you meant the [tex]e^{-3t}[/tex] to multiply the whole thing in yoru last line.
 
  • #3
986
9
What was the answer in the back of the book that you are finding conflict with? Looks right to me assuming you meant the [tex]e^{-3t}[/tex] to multiply the whole thing in yoru last line.
y=C1e-3tcos(2t) + C2e-3tsin(2t)

There's probably some trig identity eluding me :confused:
 
  • #4
45
0
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.
 

Related Threads on Finding the solution to a DE with complex roots

Replies
2
Views
2K
Replies
5
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
23
Views
2K
  • Last Post
Replies
2
Views
3K
Replies
3
Views
2K
  • Last Post
Replies
3
Views
1K
Replies
13
Views
3K
Replies
3
Views
809
  • Last Post
Replies
11
Views
3K
Top