Particular solution for dif eq

[UrbanXrisis]

$$-3y''-2y'+y=-t^2+2t+2e^{-4t}$$

i am to find the particular solution to this.

i started with the non-exponential:
$$y=At^2+Bt$$
$$y'=2At+B$$
$$y''=2A$$

$$(-6A-2B)+(-4A+B)t+A(t^2)$$
$$-6A-2B=0,-4A+B=2,A(t^2)=-1$$
$$A=-1, B=3$$

i started with the exponential:
$$y=Ce^{-4y}$$
$$y'=-4Ce^{-4t}$$
$$y''=16Ce^{-4t}$$

$$-48Ce^{-4t}+8Ce^{-4t}+Ce^{-4t}=2Ce^{-4t}$$
$$C=-\frac{2}{39}$$

so I think the particular solution is:
$$y_p=-t^2+3t- \frac{2}{39}e^{-4t}$$

but this is wrong, not sure why, any ideas?

 

[Fermat]

you have -4A + B = 2, but that is inconsistent with A = -1 and B = 3.

What have you missed?

(Check a few lines before)

 

[UrbanXrisis]

if A=-1, then b=-2? but for the other equation it is 3? i don't understand this

 

[Fermat]

You forgot the C!

 

[UrbanXrisis]

couldnt I do it separated? solve for A and B, and then solve for C would be the same as solving for A B and C at the same time right?

 

[Fermat]

y=At² + Bt + C

 

[Fermat]

My apologies Urban...
I only read your post as far as the ommission of C from the 1st eqn.
I just now saw the C from the 2nd part you did.
I must have been confusing you.
Sorry about that.

 

[UrbanXrisis]

it's no problem Fermat, any ideas is to what I have done wrong? since B can be both equal to -2 and 3, i am not sure what to do.

 

[Fermat]

do what you did before , but with y=At² + Bt + C.

y=At² + Bt + C
y' = 2At + B
y'' = 2A

-3y'' - 2y + y = -t² + t
-6A -4At - 2B + At² + Bt + C = -t² +t

etc.