# Initial value problem differentials, close to getting answer

1. Jun 25, 2008

### itunescape

1. The problem statement, all variables and given/known data
y" + 2y' + 10y=0
y(0)=1
y'(0)= -1

solve initial value problem
2. Relevant equations

e^( ~ + iu)t= e^ ~t (cos ut + i sin ut)

3. The attempt at a solution

i've gotten pretty far into the problem, but i just can't seem to get the correct final answer.

I changed y" + 2y' + 10y=0 int r^2 +2r +10= 0 and used the quadratic formula to get:

r= 1-3i and r= 1+ 3i

so for the general solution i got:
yg= y1 + y2
yg= c1e^t( cos 3t + i sin 3t) + c2e^t( cos 3t - i sin 3t)

i need to solve for the constants:

y= c1 e^t cos 3t + c2 i sin3t ( an example in the book followed this method but i dont know why there is a c2 next to sin 3t and why there isn't a y2 )

i got c1= 1 and took the derivative to find c2 and got c2= -2/3

y'= c1 e^t cos 3t -3c1e^t sin 3t + c2e^tsin3t + 3 cos3tc2e^t

overall i got yg= 2e^tcos 3t but the answer is suppose to be yg= e^-t cos 3t. why? the textbook i use has this annoying ability of skipping 90% of the arithmatic work within examples, so i'm not sure where I could have gone wrong, can u guys help? :/

2. Jun 25, 2008

### rock.freak667

remember the quadratic eq'n formula is

$$r=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

you just forgot the '-b'