# 2nd order homogeneous differentials equationsWatch

#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.
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8 years ago
#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
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#3
Ok, I'll try that, thank you for your help.
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