Finding the general solution to a differential equation

clarineterr
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Homework Statement


\frac{d^{2}y}{dt} +4\frac{dy}{dt}+20y=e^{-2t}(sin4t+cos4t)

Homework Equations





The Attempt at a Solution



The solution to the homogeneous equation: \frac{d^{2}y}{dt} +4\frac{dy}{dt}+20y=0 is

y= k1e^{-2t}cos4t +k2e^{-2t}sin4t

Then I guessed ae^{-2+4i} as a possible solution and it didn't work, and that's where I'm stuck.
 
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Since the complementary solution yc= e-2t(Acos(4t) + Bsin(4t)) is repeated in the non-homogeneous term, for your particular solution try

yp = te-2t(Ccos(4t) + Dsin(4t))
 
Nope...didn't work. :(
 
clarineterr said:
Nope...didn't work. :(

Yes, it does work. Check your work or show it here if you can't find your error.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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