Solving Particular Integral of Polynomial & Exponential

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Homework Help Overview

The discussion revolves around finding the particular integral (PI) of a differential equation involving both polynomial and exponential terms on the right-hand side (RHS). The equation presented is y'' + 5y' + 4y = x^2 + 2e^(-x), with auxiliary roots identified as m1 = -1 and m2 = -4.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the method of finding the particular integral when the RHS includes an exponential function that is also a solution to the homogeneous equation. There is discussion about the necessity of modifying the trial solution by multiplying by x due to the presence of the exponential term.

Discussion Status

Some participants have offered guidance on how to approach the problem, suggesting the inclusion of a polynomial alongside the modified exponential term in the trial solution. There appears to be some confusion regarding the combination of terms and the correct form of the particular integral.

Contextual Notes

Participants are navigating the rules of finding particular integrals in the context of differential equations, particularly when faced with the challenge of combining polynomial and exponential components. The original poster expresses uncertainty about how to proceed given the constraints of the problem.

enosthapa
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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.
 
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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:
 
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:/
 
i meant a polynomial plus a constant times xe-x

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

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

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