# Problem with general solution to a complex roots diffy Q

## 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($$\sqrt{3}$$*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

## The Attempt at a Solution

LCKurtz
Homework Helper
Gold Member
Your work is correct for the equation given. Is there supposed to be a non-homogeneous term on the right side?

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))

LCKurtz
Homework Helper
Gold Member
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).

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

Ok so apparently it accepted e^(-x/2) but not e^(-.5x). Thank you so much for your help!