View Full Version : quick help with ODE
JaysFan31
May18-07, 03:54 PM
Hi.
I have this problem:
Use the method of undetermined coefficients to find one solution of
y'' + 2y' - 7y = [(-1t^2)+4t+7]e^(2t).
Find a specific solution, not a general one.
I'm having lots of trouble finding the coefficients.
I'm assuming I need to use [(At^2)+Bt+C]*e^(2t).
This however does not seem to work. Can anyone confirm this as the function I need to use for differentiation?
Thank you.
That's the form alright - and I have no problems getting the coefficients. So try again.
JaysFan31
May18-07, 04:29 PM
I have [(At^2) + Bt + C]e^2t
Y(t) = [(At^2) + Bt + C]e^2t
Y'(t) = (2At + B)e^2t + [(At^2) + Bt + C]2e^2t
Y''(t) = (2A)e^2t + [(At^2) + Bt + C]4e^2t
Then [(At^2) + Bt + C]-7e^2t + (2At + B)2e^2t + [(At^2) + Bt + C]4e^2t + (2A)e^2t + [(At^2) + Bt + C]4e^2t = the right side of the ODE
[(At^2) + Bt + C] + 2(2At + B) + 2A = (-t^2) + 4t +7
I get that A = -1, B = 8, and C = -7. However, this is incorrect. What's wrong?
Dick, thanks for the response by the way.
Y''(t) is wrong. You didn't use the product rule enough - Y' has two products, so you should wind up with 4 terms. You are missing a (2At+B)2e^2t and a (2At+B)2e^2t.
JaysFan31
May18-07, 05:06 PM
So I get,
[(At^2) + Bt + C] + 6(At + B) + 2A = -1t^2 + 4t + 7
With this, A = -1, B = 10, C = -51.
My program still tells me this is wrong.
It still is. Maybe you could correct your previously posted detailed solution and we could look for problems. There is nothing wrong with your approach - must just be dropping some terms.
JaysFan31
May18-07, 05:39 PM
Are you sure it's not (At^2 + Bt + C)De^(2x)?
JaysFan31
May18-07, 05:48 PM
Yeah I got it allowing for a D. Not quite sure if it made a difference or not or if my arithmetic just improved, but thank you very much for the help anyway.
HallsofIvy
May18-07, 06:14 PM
Are you sure it's not (At^2 + Bt + C)De^(2x)?
That wouldn't make much sense! It's the same as (ADt2+BDt+ CD)e2x and since A, B, C, D are all unknown constants, it is no different than (At2+ Bt+ C)e2x.
D?????????????????????? I'm really curious what are the value of A,B,C and D??? turned out to be. As A=-1 was correct, D must be 1. I think the hypothesis that your arithmetic just improved is the correct one.
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