Solving $y''+2y'+y = 2e^{-t}$ by the method of undetermined coefficients I need to solve
$$y''+2'y+y = 2e^{-t},$$
using the method of undetermined coefficients (and by founding a solution for the homogeneous equation).
I tried first guessing a solution of the form $y = Ae^{-t}$, but when I tried to solve for $A$, I got a surprise: I couldn't equate the terms when I plugged $y$ in the differential equation, because I got $e^{-t}(A-2A+2A) = 2e^{-t}$ but the $A$'s sum to $0$. I then searched my book and realized that the problem was because $Ae^{-t}$ is already a solution for the homogeneous equation. I then tried $Ate^{-t}$ because my book tries it for a different equation and it worked, but in my case I got the $A$'s summing to $0$ again.
What should be my guess?
 A: Homogeneous equation:
$y''+2y'+y=0$,   $x^2+2x+1=(x+1)^2$ solution $ae^{-t}$
$y(t)=a(t)e^{-t}$, $y'(t)=a'(t)e^{-t}-a(t)e^{-t}$, $y''=a''(t)e^{-t}-a'(t)e^{-t}-a'(t)e^{-t}+a(t)e^{-t}$,
$y''2y'+y=a"(t)e^{-t}=2e^{-t}$, $a''(t)=2, a(t)=t^2+at+b, y(t)=(t^2+at+b)e^{-t}$.
A: The other answers did not address why one can use the auxiliary equation to find the complementary solution.
$$\begin{array}{rcl}
y'' + 2y' + y &=& 0 \\
y'' + y' + y' + y &=& 0 \\
e^t(y'' + y') + e^t(y' + y) &=& 0 \\
(e^t y')' + (e^t y)' &=& 0 \\
e^t y' + e^t y &=& A \\
(e^t y)' &=& A \\
e^t y &=& A t + B \\
y &=& A t e^{-t} + B e^{-t} \\
\end{array}$$
Using the same method, one doesn't actually need to use the method of undetermined coefficients:
$$\begin{array}{rcl}
y'' + 2y' + y &=& 2e^{-t} \\
y'' + y' + y' + y &=& 2e^{-t} \\
e^t(y'' + y') + e^t(y' + y) &=& 2 \\
(e^t y')' + (e^t y)' &=& 2 \\
e^t y' + e^t y &=& 2t + A \\
(e^t y)' &=& 2t + A \\
e^t y &=& t^2 + A t + B \\
y &=& t^2 e^{-t} + A t e^{-t} + B e^{-t} \\
\end{array}$$
You could never have guessed the particular solution $y=t^2e^{-t}$ using the undetermined coefficients method.
Verification:
$$\begin{array}{rcl}
y &=& t^2 e^{-t} + A t e^{-t} + B e^{-t} \\
y' &=& (2t - t^2) e^{-t} + A (1 - t) e^{-t} - B e^{-t} \\
&=& - t^2 e^{-t} + (2 - A) t e^{-t} + (A - B) e^{-t} \\
y'' &=& - (2t - t^2) e^{-t} + (2 - A) (1 - t) e^{-t} - (A - B) e^{-t} \\
&=& t^2 e^{-t} + (A - 4) t e^{-t} + (2 + B - 2A) e^{-t} \\
y'' + 2y' + y &=& [1 + 2(-1) + 1] t^2 e^{-t} + [(A - 4) + 2(2 - A) + A] t e^{-t} + [(2 + B - 2A) + 2(A - B) + B] e^{-t} \\
&=& 2 e^{-t}
\end{array}$$
Conclusion... don't use the method of undetermined coefficients?
A: Some of the answers here beat around the bush. You're supposed to multiply the particular solution by x raised to the multiplicity of "b" in a form $e^{bt}$. -1 is a double root with a multiplicity of 2 so your potential solution is multiplied by $x^2$.
In other words, the form of the particular solution is $x^2(A)e^{-t}$
I apologize for rehasing information but these answers beat around the bush. They make the method sound trivial like guesswork. In reality, the particular solution is a very specific formula. Also, your A's dont sum to 0, but you get junk nonetheless.
A: Hint: Multiply both sides by $e^t$ and then set $z = ye^t$ and express the equation in terms of $z$. 
A: It happened because $te^{-t}$ is also a solution of the homogeneous equation. Try $At^2e^{-t}$ instead.
A: We must see the complementary solution before assuming the particular solution, and the reason is to prevent similarities which may occur between the two.  For the question, the complementary solution is found when the D.E is homogeneous 
$$y''+2y'+y=0$$
$$r^2+2r+1=0$$
$$r_{1,2}=-1$$
so, the complementary solution will be
$$y_c=c_1e^{-t}+c_2e^{-t}$$
to prevent the similarity between them, we should multiply one of them by $t$, so the solution will become
$$y_c=c_1e^{-t}+c_2te^{-t}$$
the next step is to find the particular solution depending on the R.H.S which was $(e^{-t})$
If we assume the particular solution as $y_p=Ae^{-t}$, we will find this solution is found in complementary solution therefore we should multiply by $t$. The last thing is not enough  because the solution $te^{-t}$ is also found in the complementary solution, so we should multiply by $t^2$
hence the assumed particular solution must be
$$y_p=At^2e^{-t}$$
