Nonhomogeneous diff equations method of undeterined coeff.

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To solve the differential equation y'' - 2y' - 3y = 3te^-t using the method of undetermined coefficients, the characteristic equation yields roots r = -1 and r = 3, leading to a complementary solution y = c1e^-t + c2e^3t. Since e^-t is already a solution, the particular solution yp must be multiplied by t, resulting in yp = (At^2 + Bt)e^-t. The discussion emphasizes the need to include polynomial terms up to the degree of the non-homogeneous part, and it clarifies that the constant term "C" is unnecessary as it cancels out. The final solution requires careful substitution to ensure accuracy in deriving yp.
iamtrojan3
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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!
 
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iamtrojan3 said:

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.
 
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
 
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.)

iamtrojan3 said:
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
 
Thanks a lot HallsofIvy, it seems i just subbed in wrong for y''p and the "c" ended up canceling out anyways.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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