Determine the form of a particular solution to

y″+24y′+144y=t2e−12t+e−12t

Question

Determine the form of a particular solution to

y″+24y′+144y=t2e−12t+e−12t

BlackTom AI · Accepted Answer

Question restatement: Determine the form of a particular solution to the differential equation y'' + 24 y' + 144 y = t^2 e^{-12 t} + e^{-12 t}.

Option under consideration: yp = t^2 (A2 t^2 + A1 t + A0) e^{-12 t}.

Step-by-step analysis:
- First, identify the homogeneous part. The characteristic equation is r^2 + 24 r + 144 = 0, which factors as (r + 12)^2 = 0. Hence the homogeneous solution involves e^{-12 t} and t e^{-12 t}, i.e., y_h = (C1 + C2 t) e^{-12 t}.
- Next, examine the forcing terms on the right-hand side: t^2 e^{-12 t} and e^{-12 t}. Both terms contain the exponential factor e^{-12 t} with α = -12, which is a root of the characteristic equation of multiplicity 2.
- According to the method of undetermined coefficients for a forcing term of the form P_m(t) e^{α t}, where α is a root of the characteristic equation of multiplicity s, the appropriate trial solution must be t^s times a polynomial Q_m(t) of the same degree m as P_m(t), all multiplied by e^{α t}.
- Here, we have two parts to the forcing:
  1) For P_m of degree 2 (the t^2 e^{-12 t} term) with α = -12 of multiplicity s = 2, the trial is t^2 times a degree-2 polynomial: t^2 (A2 t^2 + A1 t + A0) e^{-12 t}.
  2) For P_m of degree 0 (the e^{-12 t} term) with α = -12 of multiplicity s = 2, the trial would be t^2 times a degree-0 polynomial (a constant) times e^{-12 t}, which is already encompassed by the same general form if we allow all coefficients to be determined in the degree-2 polynomial. In practice, the combined forcing can be accommodated by the same form yp = t^2 (A2 t^2 + A1 t + A0) e^{-12 t}, which yields enough freedom (A2, A1, A0) to match both parts of the forcing.
- Therefore, the proposed form yp = t^2 (A2 t^2 + A1 t + A0) e^{-12 t} is consistent with the method of undetermined coefficients for this problem, given the repeated root at r = -12 and the polynomial degrees involved in the forcing terms.
- Why other potential forms would be problematic: using a trial without the extra t^2 factor would clash with the homogeneous solutions, since e^{-12 t} and t e^{-12 t} are already solutions of the homogeneous equation; similarly, an insufficient polynomial degree would fail to provide enough degrees of freedom to satisfy both forcing terms upon substitution.
- In summary, the form yp = t^2 (A2 t^2 + A1 t + A0) e^{-12 t} is appropriate because it compensates for the repeated root and accommodates a polynomial forcing of degree 2, giving a flexible enough Ansatz to determine the particular solution.

MAP2302 L15 section 4.5

Determine the form of a particular solution to y″+24y′+144y=t2e−12t+e−12t

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