2nd Order In-Homogeneous Particular Solution?

raaznar
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Yes, the characteristic equation is r^2+ 4r+ 3= (r+ 3)(r+ 1)= 0 which has roots r= -1 and r= -3 so the general solution to the associated homogeneous equation is Ae^{-3x}+ Be^{-x}.

For 2e^{2x}cos(x) try e^{2x}(Ccos(x)+ Dsin(x). For ##3xe^{-4x}##, try e^{-4x}(Ex+ F). For 3, try G.
 
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Thread 'Finding the nth roots of a complex number'

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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