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# 2nd order homogeneous differentials equations watch

1. Hi, I have been asked to solve the following equation:
y'' + 2y' +5y = 0 with the initial conditions y(0) = -1, y'(0) = 1+2*sqrt(3)
In the form Y(x) = A*e^kx*cos((omega)x - (phi)) Where A is a constant, k is the real root of the auxillary equation, and (omega) is the imaginary part of the root of the auxillary equation.
Thus far I have solved the axillary equation, used Eulers formula and worked out the necessary constants from the given conditions to get the equation in the form:
y(x) = e^(-x)*(-cos(2x)+sqrt(3)sin(2x))

I just wondered if anyone could give me any pointers on how to get from this form of the solution to the one required. Thanks in advance, Kam.
2. if what you have got so far is correct:

compare -cos(2x)+rt(3)sin(2x) with Rcos(2x-phi)=Rcos(phi)cos(2x)+Rsin(phi)s in(2x)

comparing gives values for Rcos(phi) and Rsin(phi).which can be used to find R and phi
3. Ok, I'll try that, thank you for your help.

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Updated: January 27, 2010
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