- #1
EvLer
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Hello,
I have this DE:
y'' + 2y' - 3y = 8ex - 12e3x
when I find homogeneous solution I get
yh = c1ex + c2e-3x;
so now to find the particular solution by method of undetermined coefficients, do I set y to smth like this:
y = y1 + y2
where
y1 = Axex,
y2 = A3x ?
since one of the solutions to auxiliary equation appears on the RHS of the DE and the other does not?
I don't need the full solution, just confirmation/correction of this part.
Thanks much!
EDIT: if I take the fact that if y1 + y2 is a solution, then y1 is a solution and y2 is a solution. I guess I answered my own question.![Frown :frown: :frown:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
I have this DE:
y'' + 2y' - 3y = 8ex - 12e3x
when I find homogeneous solution I get
yh = c1ex + c2e-3x;
so now to find the particular solution by method of undetermined coefficients, do I set y to smth like this:
y = y1 + y2
where
y1 = Axex,
y2 = A3x ?
since one of the solutions to auxiliary equation appears on the RHS of the DE and the other does not?
I don't need the full solution, just confirmation/correction of this part.
Thanks much!
EDIT: if I take the fact that if y1 + y2 is a solution, then y1 is a solution and y2 is a solution. I guess I answered my own question.
Last edited: