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Answers for “该片段在待定系数法中停在哪里?”

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

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

Meet the concept

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

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

Meet the concept

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

Conditions: The ODE is x¨+3x˙+2x=2e−t\ddot{x}+3\dot{x}+2x=2e^{-t}.; The coefficient AA has been determined to be 22.; The trial solution form is x=Ate−tx=Ate^{-t}.

Know when to use it

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

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