Nonhomogeneous diff equations method of undeterined coeff.

[iamtrojan3]

Homework Statement

Find the general solution to the diff equation using undetermined coefficients
y''-2y'-3y = 3te^-1

Homework Equations

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!

 

[Mark44]

Homework Statement

Find the general solution to the diff equation using undetermined coefficients
y''-2y'-3y = 3te^-1

Homework Equations

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!

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.

 

[iamtrojan3]

thanks for responding, Let's 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

 

[HallsofIvy]

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.)

thanks for responding, Let's 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

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

 

[iamtrojan3]

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