# Nonhomogeneous diff equations method of undeterined coeff.

1. Oct 20, 2009

### iamtrojan3

1. The problem statement, all variables and given/known data
Find the general solution to the diff equation using undetermined coefficients
y''-2y'-3y = 3te^-1

2. Relevant equations

3. The attempt at a solution
r^2 - 2r -3 = 0
r = -1, 3
so y = c1 e^-t + c2e^3t + yp
since e^-t already exists as a solution, i have to multiply my Yp by t to make sure i don't' end up with the same solution.
So my Yp with the unknown coeffcient should be,
yp = At^2e^-t
should it be (At^2+bt+c)e^-t instead or something else? i've done it both ways and came up with wrong answers.
I know what to do after figuring out yp, its just i can't get it right with ideas i have right now.
Any help is greatly appreciated.
Thank you!

2. Oct 20, 2009

### Staff: Mentor

Yes, go with (At2 + Bt + C)e-t for your particular solution. Show us your work, and we'll figure out what's going wrong.

3. Oct 20, 2009

### iamtrojan3

thanks for responding, Lets see if i have the yp's correct
yp = (At^2+Bt+c)e^-t
y'p = (2At + B)e^-t -(At^2+Bt+c)e^-t
y''p = 2Ae^-t - (2At + B)e^-t - (2At+B)e^-t -(At^2+Bt+c)e^-t

4. Oct 20, 2009

### HallsofIvy

Staff Emeritus
In general, it you have a power of t on the right side, you will need to try a polynomial up to that power. If your right side were $te^{at}$ and $e^{at}$ were not already a solution, you would try $(At+ B)e^{at}$. Since, here, $e^{at}$ is a solution, multiply that by t: try $(At^2+ Bt)e^{at}$.

You should find that you do NOT need that "c". (Using it will just give C= 0.)

You can add those two middle terms and the last term should be "+".
y"/= 2Ae^-t- 2(2At+ B)e^-t+ (At^2+ Bt+ C)e^-t

5. Oct 20, 2009

### iamtrojan3

Thanks alot HallsofIvy, it seems i just subbed in wrong for y''p and the "c" ended up canceling out anyways.