# Auxiliary Eqns with Complex Roots

1. Oct 3, 2009

### marco101

1. The problem statement, all variables and given/known data
solve the initial value problem
y''+9y = 0, y(0) = 1, y'(0) = 1

2. Relevant equations
gen solution form is y(t) = C1e^At*(cosBt) + C2e^At*(sinBt)
where A is the real number and B is the imaginary number

3. The attempt at a solution
i just wanted to check if im doing this right.
i started off finding the roots,
r^2 + 9
r = +/- 3i
since the real is 0, e would be raised to 0 and become one leaving me with C1cos3t + C2sin3t

2. Oct 3, 2009

### rock.freak667

Yes that is correct. Now find y(0) and y'(0)

3. Oct 4, 2009

### marco101

since there are double roots, should i make changes to the general solution form? everything cancels out to be zero because y'' = -9cos3t -9sin3t and y = 9cos3t + 9sin3t because it was multiplied by 9 in the original equation.

4. Oct 4, 2009

### rock.freak667

double roots would mean r=3i,3i or r=-3i,-3i. You have r=0±3i. So yes your general solution is correct.

Also from y = 9cos3t + 9sin3t, y''≠ y'' = -9cos3t -9sin3t , recheck your algebra on that one.

Anyway you take it y= 9cos3t +9sin3t is correct.