# Homework Help: Quick help with ODE

1. May 18, 2007

### JaysFan31

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.

2. May 18, 2007

### Dick

That's the form alright - and I have no problems getting the coefficients. So try again.

3. May 18, 2007

### JaysFan31

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.

4. May 18, 2007

### Dick

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.

Last edited: May 18, 2007
5. May 18, 2007

### JaysFan31

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.

6. May 18, 2007

### Dick

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.

7. May 18, 2007

### JaysFan31

Are you sure it's not (At^2 + Bt + C)De^(2x)?

8. May 18, 2007

### JaysFan31

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.

9. May 18, 2007

### HallsofIvy

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.

10. May 18, 2007

### Dick

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.

Last edited: May 18, 2007