## anonymous 4 years ago I have a question on complex roots. I have the eqn 8y'' +18y' +18y = 0, with the initial conditions of y(0)=3 and y'(0) = 4. I am trying to put this in the form y = Ae^rx +Be^rx, with my r = -9/8 +/- sqrt(126)....can anyone give pointers on getting the correct A and B? My answers are still wrong :(

1. anonymous

Hello!

2. JamesJ

Remember that $\sin bx = \frac{1}{2i} ( e^{bxi} - e^{-bxi})$ and $\cos bx = \frac{1}{2} ( e^{bxi} + e^{-bxi})$ So you can write your solutions as the product of a (real) exponential and of trig functions.

3. JamesJ

In particular, if you find the roots to the characteristic equation are $\lambda = a \pm bi$, then the general solution to the homogenous 2nd order equation is $C_1 e^{ax} \cos bx + C_2 e^{ax} \sin bx$

4. anonymous

Oh, okay. So when I know I will have complex numbers as a result of solving for my roots, how should I go about identifying where I should use this rule?

5. JamesJ

whether you have roots which have a non-zero imaginary part

6. JamesJ
7. anonymous

Thanks so much!