Where does the first segment stop in the method of undetermined coefficients for the ODE $\ddot{x}+3\dot{x}+2x=2e^{-t}$?

The particular solution uses $Ate^{-t}$ because the forcing term $e^{-t}$ is already a solution to the associated homogeneous equation. In the method of undetermined coefficients, if the forcing function overlaps with a homogeneous solution (resonance), the standard guess must be multiplied by $t$ to ensure linear independence.

Conditions: The ODE is linear with constant coefficients.; The forcing term contains $e^{-t}$.; $r=-1$ is a root of the homogeneous characteristic equation.

Substitute the trial solution $x=Ate^{-t}$ and its derivatives into the ODE, factor out the common term $e^{-t}$, and collect like terms by powers of $t$. Equating the remaining constant term to the right-hand side yields $A=2$.

Conditions: The trial solution is $x=Ate^{-t}$.; The ODE is $\ddot{x}+3\dot{x}+2x=2e^{-t}$.; $e^{-t}$ is never zero, allowing it to be canceled from both sides.

The first derivative is $\dot{x}=Ae^{-t}(1-t)$, and the second derivative is $\ddot{x}=Ae^{-t}(t-2)$. These are computed by applying the product rule successively to the trial solution $x=Ate^{-t}$.

Conditions: The trial solution is $x=Ate^{-t}$.; Differentiation is with respect to $t$.; The product rule is applied to $t$ and $e^{-t}$.

The final particular solution is $X_p(t) = 2te^{-t}$. This is obtained by substituting the solved coefficient $A=2$ back into the resonant trial solution $x=Ate^{-t}$.

Conditions: The ODE is $\ddot{x}+3\dot{x}+2x=2e^{-t}$.; The coefficient $A$ has been determined to be $2$.; The trial solution form is $x=Ate^{-t}$.