# 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

Last edited:

Homework Helper
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