# Particular Integral

y''+5y'+4y=x^2+2e^(-x)
Auxiliary roots: m1=-1 m2=-4

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

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tiny-tim
Homework Helper
hi enosthapa!

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

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! ) …

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)

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:/

tiny-tim
Homework Helper
i meant a polynomial plus a constant times xe-x

doesn't that work?

I don't have the answer which could have been helpful. But thanks anyway.

Redbelly98
Staff Emeritus