Solving y''+y'+y=0: Two Solutions?

[tennishaha]

Homework Statement

y''+y'+y=0

Homework Equations

say we get the roots: -1/2+i*sqrt(3)/2 and -1/2-i*sqrt(3)/2

The Attempt at a Solution

I saw two solutions to this problem
first is
y=e^(-1/2*x)(c1*cos(sqrt(3)/2*x)+c2*sin(sqrt(3)/2*x))
second is
y=c1*e^[(-1/2+i*sqrt(3)/2)*x]+c2*e^[(-1/2-i*sqrt(3)/2)*x]

which one is correct? I think the two solutions are different
thanks

 

[Raskolnikov]

Homework Statement

y''+y'+y=0

Homework Equations

say we get the roots: -1/2+i*sqrt(3)/2 and -1/2-i*sqrt(3)/2

The Attempt at a Solution

I saw two solutions to this problem
first is
y=e^(-1/2*x)(c1*cos(sqrt(3)/2*x)+c2*sin(sqrt(3)/2*x))
second is
y=c1*e^[(-1/2+i*sqrt(3)/2)*x]+c2*e^[(-1/2-i*sqrt(3)/2)*x]

which one is correct? I think the two solutions are different
thanks

Euler's Formula states:

e^{i\theta} = \cos(\theta) + i\sin(\theta).

Can you fill in the remaining steps? I hope this helps.

 

[HallsofIvy]

Using Euler's formula, as Raskolnikov suggests, you should find that the two functions are the same. (Possibly different constants, of course.)