General Solution to Non-homologous ODEs

zabumafu
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Homework Statement


Find the general solution of the given differential equation:

y''+y'+4y=2sinht


Homework Equations



I believe sinht=(e^t-e^-t)/2

The Attempt at a Solution



I tried to find the general equation if it were homogenous however I get the roots are
r=[1+- (-15)^.5]/2 and get stuck. If anyone can help me figure out the rest of the problem I should be able to teach myself how to do the rest of them. I know the answer is:

y1=C1*e^(-t/2)cos(root(15t/2))+C2*e^(-t/2)sin(root(15t/2))+1/6e^t-1/4e^-t
 
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You see how the final answer looks like and still you don't know how to answer this?

In the final answer the first two terms are the solution of the homogenous DE, and the rest two terms are the private solution, i.e you guess: y_p = Ae^t+Be^-t and then plug it to the DE, and equate the coeffiecients on both sides of the equation, such that the coeff of e^t on one side is the same on the other side, the same with e^-t, this is how you find A and B.
 
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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