Determine the form of a particular solution to

y″−y=9e5t+6te5t+3t2e5t

Question

Determine the form of a particular solution to

y″−y=9e5t+6te5t+3t2e5t

BlackTom AI · Accepted Answer

We start by identifying the forcing term on the right-hand side: a polynomial in t multiplying e^{5t}, namely 9e^{5t} + 6t e^{5t} + 3t^2 e^{5t}. The differential equation is y'' - y = ..., with characteristic equation r^2 - 1 = 0, which has roots r = ±1. Since the exponential factor in the forcing, e^{5t}, corresponds to r = 5, which is not a root of the homogeneous equation, there is no resonance to worry about. Therefore, the form of a particular solution should be a polynomial in t of degree 2 multiplied by e^{5t}, i.e., (A2 t^2 + A1 t + A0) e^{5t}.

Option-by-option analysis:
- Option 1: yp(t) = (A2 t^2 + A1 t + A0) e^{5t} + (B2 t^2 + B1 t + B0) e^{-5t}
This introduces an extra term with e^{-5t}. Since the forcing term only involves e^{5t} and not e^{-5t}, and there is no need for a complementary (homogeneous) solution term here, this form is unnecessarily broad and incorrect for a standard particular solution unless resonance or a mixed forcing required it. Therefore this option overcounts the necessary structure.
- Option 2: yp(t) = (A2 t^2 + A1 t + A0) e^{5t}
This matches the expected form: a degree-2 polynomial in t multiplied by e^{5t}, which is appropriate because r = 5 is not a root of the homogeneous equation, and the forcing term is a degree-2 polynomial times e^{5t}. This aligns with the standard method of undetermined coefficients.
- Option 3: yp(t) = t^2 (A2 t^2 + A1 t + A0) e^{5t}
Here the polynomial factor is multiplied by an extra t^2, yielding a degree-4 polynomial times e^{5t}. This is excessive given the forcing term is only degree 2 in t; such a higher-degree ansatz would be used if there were resonance or if the forcing demanded more degrees of freedom, which is not the case here.
- Option 4: yp(t) = t (A2 t^2 + A1 t + A0) e^{5t}
This option uses a polynomial of degree 3 in t overall (due to the extra factor t) but only with a single factor of t, whereas the standard approach for a degree-2 polynomial forcing typically yields a degree-2 polynomial in t multiplied by e^{5t}. This underestimates or misweights the necessary polynomial degree, making it an incorrect choice for this problem.

In summary, none of the alternatives aside from the second match the conventional undetermined-coefficients form for a right-hand side of (polynomial degree 2) times e^{5t}, given no resonance at r = 5. The second option precisely captures the required structure without introducing unnecessary or insufficient factors.

Determine the form of a particular solution to y″−y=9e5t+6te5t+3t2e5t

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

Determine the form of a particular solution to y″−y=9e5t+6te5t+3t2e5t

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