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Particular Integral

  • Thread starter enosthapa
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  • #1
9
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y''+5y'+4y=x^2+2e^(-x)
Auxiliary roots: m1=-1 m2=-4

Using trial please
cant seem to work out PI with polynomial and exponential together.
 

Answers and Replies

  • #2
tiny-tim
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hi enosthapa! :smile:

(try using the X2 icon just above the Reply box :wink:)

when your RHS contains an exponential which is a solution of the LHS, it obviously can't be a PI on its own (as you've found out! :redface:) …

you need to multiply it by x (or xn is it's an nth root of the LHS) …

in this case, try xe-x (plus a polynomial) :wink:
 
  • #3
9
0
Thanks a lot but I am still confused
As far as i know,
if there is exponential on RHS (ekx) and if 'k' is simple root of the auxiliary equation which it is in this case, then we try a.x.ek.x i.e a.x.e-x

but the question has polynomial in addition so did you mean i have to add the 'PI' I would get if there was polynomial only on RHS?
sorry i m confusing u:/
 
  • #4
tiny-tim
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i meant a polynomial plus a constant times xe-x

doesn't that work? :confused:
 
  • #5
9
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I don't have the answer which could have been helpful. But thanks anyway.
 
  • #6
Redbelly98
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Moderator's note: thread moved from "Differential Equations".

The usual rules rules for giving homework help are now in effect.
 

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