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Why does the particular solution use Ate−tAte^{-t} instead of Ae−tAe^{-t} for the ODE x¨+3x˙+2x=2e−t\ddot{x}+3\dot{x}+2x=2e^{-t}?

The particular solution uses Ate−tAte^{-t} because the forcing term e−te^{-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 tt to ensure linear independence.

Conditions

  • The ODE is linear with constant coefficients.
  • The forcing term contains e−te^{-t}.
  • r=−1r=-1 is a root of the homogeneous characteristic equation.

Reasoning, step by step

  1. Solve the homogeneous equation x¨+3x˙+2x=0\ddot{x}+3\dot{x}+2x=0 using the trial x=ertx=e^{rt}.
  2. Find the characteristic equation r2+3r+2=0r^2+3r+2=0, which factors to (r+2)(r+1)=0(r+2)(r+1)=0.
  3. Identify the homogeneous roots r=−2r=-2 and r=−1r=-1, giving the homogeneous solution xh(t)=c1e−2t+c2e−tx_h(t)=c_1e^{-2t}+c_2e^{-t}.
  4. Notice that the forcing term 2e−t2e^{-t} shares the exponential factor e−te^{-t} with the homogeneous solution.
  5. Modify the naive particular guess Ae−tAe^{-t} by multiplying by tt to avoid duplication, resulting in xp(t)=Ate−tx_p(t)=Ate^{-t}.

Example

The speaker says, 'the complication is that e−te^{-t} is a solution of the homogeneous equation... so we need to multiply now the right hand side by t, giving Ate−tA t e^{-t}.' An arrow visually links the forcing term 2e−t2e^{-t} to the e−te^{-t} term in xh(t)x_h(t).

Common misconceptions

  • Assuming the naive guess Ae−tAe^{-t} will work, which fails because it is already part of the homogeneous solution.
  • Forgetting to multiply by tt when the forcing term matches a homogeneous mode.

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