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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 :(
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 :(

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Remember 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.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1In 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 \[ C_1 e^{ax} \cos bx + C_2 e^{ax} \sin bx \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, 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?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1whether you have roots which have a nonzero imaginary part

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1I recommend you watch this lecture: http://ocw.mit.edu/courses/mathematics/1803differentialequationsspring2010/videolectures/lecture9solvingsecondorderlinearodeswithconstantcoefficients/
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