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?
Best Answer
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?