Question

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 :(

Answer 1

Hello!

Answer 2

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.

Answer 3

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
\[ C_1 e^{ax} \cos bx + C_2 e^{ax} \sin bx \]

Answer 4

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?

Answer 5

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

Answer 6

I recommend you watch this lecture:
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-9-solving-second-order-linear-odes-with-constant-coefficients/

Answer 7

Thanks so much!