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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 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 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.

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.

The homogeneous roots are obtained by solving the associated homogeneous equation $\ddot{x}+3\dot{x}+2x=0$. By substituting the exponential trial solution $x=e^{rt}$, the differential equation converts into the characteristic polynomial $r^2+3r+2=0$, which factors to $(r+2)(r+1)=0$, yielding the roots $r=-2$ and $r=-1$.

Conditions: The ODE has constant coefficients.; The trial solution is of the form $x=e^{rt}$.; The characteristic equation is a quadratic polynomial.

The homogeneous roots are obtained by solving the associated homogeneous equation $\ddot{x}+3\dot{x}+2x=0$. By substituting the exponential trial solution $x=e^{rt}$, the differential equation converts into the characteristic polynomial $r^2+3r+2=0$, which factors to $(r+2)(r+1)=0$, yielding the roots $r=-2$ and $r=-1$.

Conditions: The ODE has constant coefficients.; The trial solution is of the form $x=e^{rt}$.; The characteristic equation is a quadratic polynomial.

The first segment stops after computing the first and second derivatives of the resonant trial solution $x=Ate^{-t}$. It does not yet substitute these derivatives back into the differential equation or solve for the unknown coefficient $A$.

Conditions: The segment covers the setup of the homogeneous solution and the resonant trial solution.; The derivatives $\dot{x}$ and $\ddot{x}$ are computed but not substituted.

The first segment stops after computing the first and second derivatives of the resonant trial solution $x=Ate^{-t}$. It does not yet substitute these derivatives back into the differential equation or solve for the unknown coefficient $A$.

Conditions: The segment covers the setup of the homogeneous solution and the resonant trial solution.; The derivatives $\dot{x}$ and $\ddot{x}$ are computed but not substituted.