What is the Correct Particular Solution for y''-2y'-3y=-3te^-t?

[blizzard750]

y''-2y'-3y=-3te^-t

i know that that the general solution to this problem is
yh = c1e^3t + c2e^-t

i am having trouble figuring out what the particular solution is (yp)
i keep getting the yp = 3/4te^-t , but wolfram alpha is telling me that the answer is something else.

how do i get the right particular solution. i am using the form (at+b)e^-t.

 

[lurflurf]

What is y'p'-2yp'-3yp?

You need to consider particular solutions of the form
(a t^2+b t)e^-t=(a t+b)t e^-t.

This is easy to see as the minimal equation for the forcing function is
(D+1)^3 (-3te^-t)=0
note that we need not consider e^-t as it is a solution of the homogeneous equation.

 

[HallsofIvy]

If e^{-t} were not already a solution to the associated homogeneous equation, then you would try (At+ B)e^{-t} as a patricular solution. But because it is, if you do you will get
y'= Ae^{-t}- (At+ B)e{-t}= (-At+ A- B)e^{-t}
y''= -2Ae^{-t}+ (At+ B)e^{-t}= (At+ B- 2A)e^{-t}

y''- 2y- 3y= (At+ B- 2A-(-2At+ 2A- 2B)- (3At+ 3B)e^{-t}= (A+ 2A- 3A)te^{-t}+ (-2A+ 2A+ B+ 2B- 3B)= 0
for all[\b] A, B.

Because e^{-t} is already a solution to the homogeneous equation, we need to multiply by t: try (At^2+ Bt)e^{-t}