# Second order diff eq with i

Crap, nevermind, I left b^2 out of the quadratic formula, thanks anyways.

Here's the question:
Find y as a function of t from the diff eq:
y''+6y'+25y=0 with the initial conditions y(0)=8 and y'(0)=8

I used the form r^2+6r+25=0 to solve for r and through the quadratic equation got r = -3+/-5i
so my equation now looks like
c1*e^(-3t)*cos(5t)+c2*e^(-3t)*sin(5t)=8
The second part of the equation cancels and I'm left with c1=8
Now to find y'(0) I have:
-3*c1*e^(-3t)*cos(5t)+c1*e^(-3t)*(-5*sin(5t))+-3*c2*e^(-3t)*sin(5t)
+c2*e^(-3t)*(5*cos(5t))
The second and third terms in that cancel out and I'm left with
-3*c1*e^(-3t)*cos(5t)+c2*e^(-3t)*5*cos(5t), and plugging in 0 for t and 8 for c1 I get:
-24+5*c2=8
5*c2=32
c2=32/5

So the final equation looks like
8*e^(-3t)*cos(5t)+32/5*e^(-3t)*sin(5t)

This isn't right and I can't figure out what I did wrong. Any help would be awesome.

Thanks a lot.

## Answers and Replies

cristo
Staff Emeritus
Science Advisor
Crap, nevermind, I left b^2 out of the quadratic formula, thanks anyways.

Does this mean you've solved the problem?