2nd order DE: finding particular solution

zyferion
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Homework Statement


Find the general solution of the following differential equation:
y" + 3y' + 2y = sin ex


Homework Equations


y = yh + yp

homogeneous solution: (found by solving characteristic eq)
yh = Ae-2x + Be^-x

The Attempt at a Solution


from my table if r(x) = ksin(wx)
then choice for yp = Kcos(wx) + Msin(wx)

i tried using y_p = Kcos(ex) + Msin(ex)
and found y' and y" using chain and product rule but it ended up messy and i couldn't cancel things out in the end.
if you've come across something like this please help me find the general form?
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Are you sure the problem doesn't have a typo in it? That forcing function isn't typical of homework problems.
 
Blegh don't read off of a table. If the inhomogeneous problem was on that table you could have just as well guessed the solution immediately. I'm pretty sure this can be done via Green's functions or variation of parameters or whatever it's called, as it's the tool you turn to when you can't guess easily.
 
snipez90 said:
Blegh don't read off of a table. If the inhomogeneous problem was on that table you could have just as well guessed the solution immediately. I'm pretty sure this can be done via Green's functions or variation of parameters or whatever it's called, as it's the tool you turn to when you can't guess easily.

yeah i got it now, i used variation of parameters. For some reason i got stuck using "general form" thinking.
 
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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