Question in complex general equation (2nd order)

[batmankiller]

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(\sqrt{3}*x/2)+bsin(\sqrt{3}*x/2))

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

 

[tiny-tim]

hi batmankiller!

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

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(\sqrt{3}*x/2)+bsin(\sqrt{3}*x/2))

looks ok to me

perhaps it's your notation …

have you tried entering x/2 (instead of 0.5 x), or (√3/2)x, or 0.866 x ?

 

[batmankiller]

umm yeah thanks so much. -.5x apparently doesn't equal -x/2 anymore. Math as we know it has changed!

 

[tiny-tim]

the computers have taken over …

we have to speak the language of the conquerors! ​