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can't find the particular integral

y'' + 3y' + 2y = xe^(-x)

The complementary function is Ae(-2x) + Be(-x)

For the particular integral I tried Cxe(-x) and (Cx + D)e^(-x) and neither works. I'm at a loss.
Original post by StarvingAutist
y'' + 3y' + 2y = xe^(-x)

The complementary function is Ae(-2x) + Be(-x)

For the particular integral I tried Cxe(-x) and (Cx + D)e^(-x) and neither works. I'm at a loss.


Try: (Cx+D)(xe−x) \displaystyle (Cx+D)(xe^{-x}) for your particular integral. Since the RHS contains one of the solutions to the homogeneous ODE, that is e−x \displaystyle e^{-x} you need to multiply that term by x \displaystyle x .
(edited 8 years ago)
Original post by StarvingAutist
y'' + 3y' + 2y = xe^(-x)

The complementary function is Ae(-2x) + Be(-x)

For the particular integral I tried Cxe(-x) and (Cx + D)e^(-x) and neither works. I'm at a loss.


Have you tried Cx2e−xCx^2 e^{-x}?
Original post by atsruser
Have you tried Cx2e−xCx^2 e^{-x}?


Yeah, that didn't work either. Maybe I'm making an algebraic mistake somewhere..
Original post by StarvingAutist
Yeah, that didn't work either. Maybe I'm making an algebraic mistake somewhere..


Did you try what I suggested?
Original post by poorform
Did you try what I suggested?


Yeah, doing y = (ax^2 + bx + c)e^-x works with c unnecessary. Missed your post but thanks, you were right.
Original post by StarvingAutist
Yeah, doing y = (ax^2 + bx + c)e^-x works with c unnecessary. Missed your post but thanks, you were right.


No problem, and multiplying by the e^(-x) by x is equivalent to multiplying the (Cx+D) bit by x which is what you did in your post.
Original post by poorform
No problem, and multiplying by the e^(-x) by x is equivalent to multiplying the (Cx+D) bit by x which is what you did in your post.


Can you help with another one? I have x'' + 4x = t + cos2t; I thought x = at + b + csin2t + dcos2t would work but that just comes out as 4at + 4b = t + cos2t.
Reply 8
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TeeEm
19
Original post by StarvingAutist
Can you help with another one? I have x'' + 4x = t + cos2t; I thought x = at + b + csin2t + dcos2t would work but that just comes out as 4at + 4b = t + cos2t.


the "trig bit" in this one needs tsin2t ONLY
(edited 8 years ago)
Original post by TeeEm
the "trig bit" in this one needs tsin2t ONLY


Okay, thanks :smile:
Can you explain why?
Reply 10
Avatar for TeeEm
TeeEm
19
Original post by StarvingAutist
Okay, thanks :smile:
Can you explain why?


Experience is the answer.

For SHM differential equations with forcing

with RHS sinax you try y=sinax
with RHS cosax you try y=cosax
with RHS sinax you try y=xcosax if sinax is part of the CF
with RHS cosax you try y=xsinax if cosax is part of the CF
(edited 8 years ago)
Reply 11
Avatar for TeeEm
TeeEm
19
Original post by StarvingAutist
Okay, thanks :smile:
Can you explain why?


sorry EDIT tsin2t
(edited 8 years ago)

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