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Question in complex general equation (2nd order)

  • #1
Question:
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.)

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
 
Last edited:

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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hi batmankiller! :smile:

(have a square-root: √ and try using the X2 icon just above the Reply box :wink:)
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.)

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

perhaps it's your notation …

have you tried entering x/2 (instead of 0.5 x), or (√3/2)x, or 0.866 x ?
 
  • #3
umm yeah thanks so much. -.5x apparently doesn't equal -x/2 anymore. Math as we know it has changed!
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
250
the computers have taken over :redface:

we have to speak the language of the conquerors! :smile:
 

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