Complex exponentials and differential equations

1. Oct 23, 2005

Pixter

question: Use complex exponentials to find the solution of the differential equation

(d^2y(t)/dt^2) + (3dy(t)/dt) + (25/4)y(t) = 0

such that y(0) = 0, dy/dt =1 for t=o

my taughts: I started by putting it in the form m^2 + 3m +25/4
m = (-3sqrt(9-25))/2 = (-3sqrt(-16))/2 = (-3+-4i)/2

then i thaught one can put it in the form e^pt(AcosQt+BsinQt) [p+-Qi]

so: y = (e^(-3/2)t)(Acos2t + Bsin2t)

y(0)=0 dy/dt=1 for t=0 y=Ae^(((-3+4i)/2)t) + Be^(((-3-4i)/2)t)

0 = (e^(-3/2)t)(Acos2t + Bsin2t)
0 = (Acos2t + Bsin2t)
0 = A + 0
A=0
dy/dt = (-3/2(e^(-3/2)t))(Acos2theta + Bsin2theta) + (e^(-3/2)t)(-2Asin2t + 2Bcos2t)
1=(-3/2)A +2B
2B = 1 (because A=0)
B=1/2
so y(t) = e^((-3/2)t) ((1/2)sin2t)

Don't know if I have done the question right or even got the question at all. just wanted to know if this is right, or if i'm on the right track but made a misstake on the way. Also if I'm completly wrong please point that out and give me a pointer where to start.

edit: sorry for posting this in both precalc and in calc, just didn't know where it belongs....i'm swedish don't actually know the definition for calculus.

2. Oct 23, 2005

Galileo

The nice thing about differential equations is that you can always check if your answer is correct. Just plug your answer into the equation and see if it works out. You can also check that it does satisfy the boundray conditions.

The method was correct too. I`m not even sure why you're so unsure about the question or your answer. It asks to find the solution to the D.E. satisfying the boundary conditions and you have. Good job.