# Thread: Method of Undetermined Coefficients

1. ## Method of Undetermined Coefficients

$y"+9y = t^2(e^{3t}) + 6$

ok wow, I have racked my brain at on this for about 5 hours with and without a study group.

I cannot seem how to set up the initial part of finding the particular solution.

I had found one variable and one point which was c = 1/162 but couldnt find A, or B.

I tried using $y(t)=(e^{3t})(at^{2} + bt +c)$ but got stock when plugging in the second derivative in and couldnt not find answer for the rest of the coefficients.

what am I doing wrong?

2. Hello, p00ndawg!

$y"+9y \:=\: t^2e^{3t} + 6$

I cannot seem how to set up the initial part of finding the particular solution.

I tried: . $y(t)\:=\:e^{3t}\left(At^2 + Bt +C\right)$ but got stuck.
We need something to create the final "6".

I tried: . $y \:=\:At^2e^{3t} + Bte^{3t} + Ce^{3t} + D$

And came up with: . $A = \frac{1}{18},\;B = -\frac{1}{27},\;C = \frac{1}{162},\;D = \frac{2}{3}$

3. Originally Posted by Soroban
Hello, p00ndawg!

We need something to create the final "6".

I tried: . $y \:=\:At^2e^{3t} + Bte^{3t} + Ce^{3t} + D$

And came up with: . $A = \frac{1}{18},\;B = -\frac{1}{27},\;C = \frac{1}{162},\;D = \frac{2}{3}$

wait im not understanding, what did you do?

that is right tho, because b should equal -6 when you factor out the 162.

what i did was right? or wrong?