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Problem with general solution to a complex roots diffy Q

  • #1

Homework Statement




Question: I'm confused about how to app
Find the real-valued general solution of the differential equation

y''+1y'+1y=0
where primes indicate differentiation with respect to x. (Use the parameters a, b, etc., for undetermined constants in your solution.)

The Attempt at a Solution


My attempt:
Use characteristic equation:
r^2+r+1=0
I used the quadratic formula and got r=(-1 +-sqrt(-3))/2
So we get a=-.5 b-sqrt(3)/2

Following e^(ax)(c1cos(bx)+c2sin(bx):

we get e^(-.5x)(a*cos([tex]\sqrt{3}[/tex]*x/2)+bsin([tex]\sqrt{3}[/tex]*x/2))

Anyone see any flaw i nlogic of algebra or math? I can't seem to get a correct answer

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
LCKurtz
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Your work is correct for the equation given. Is there supposed to be a non-homogeneous term on the right side?
 
  • #3
Nope, it's given as a homogeneous equation (set equal to 0). Unless you're saying there's something i'm missing?
The full equation is y=e^(-.5x)(a*cos(*x/2)+bsin(*x/2))
 
  • #4
LCKurtz
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Nope, it's given as a homogeneous equation (set equal to 0). Unless you're saying there's something i'm missing?
The full equation is y=e^(-.5x)(a*cos(*x/2)+bsin(*x/2))
What led you to believe it was wrong? (Don't forget the sqrt(3)'s).
 
  • #5
Eh it's online homework and it says "answer is incorrect" -.-"
I entered:
e^(-.5x)(a*cos(sqrt(3)*x/2)+bsin(sqrt(3)*x/2)) and my previewed answer was in the correct format too
 
  • #6
Ok so apparently it accepted e^(-x/2) but not e^(-.5x). Thank you so much for your help!
 

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