We proceed by considering each term at the right separately
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First notice that the characteristic polynomial of the differential equation is . Hence, both and are solutions of the differential equation. If we consider the solution to be by a second order polynomial we have
then after inserting in the differential the first two terms cancel out. So they can be arbitrary since they can be canceled out with the homogeneous part.
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For this case we suppose a solution of the form
since the nonhomogeneous term is the trigonometric term .
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Since the term is a zero degree polynomial (i.e. a constant) we suppose the solution to be a polynomial of the same degree
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Finally we construct the whole particular solution by adding the terms
and if one wish to take ,
.
Use the method of undetermined coefficients to find a suitable form for the particular solution of...
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