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 restating the problem: Determine the form of a particular solution for y'' − y = 9e^{5t} + 6t e^{5t} + 3t^2 e^{5t}.

First, examine the forcing term on the right-hand side. It consists of terms that are e^{5t} multiplied by polynomials in t, with the highest polynomial degree being 2 (from 3t^2 e^{5t}).

Next, consider the homogeneous equation y'' − y = 0. Its characteristic equation r^2 − 1 = 0 has roots r = ±1, so e^{5t} is not a solution to the homogeneous equation. Therefore, there is no resonance with e^{5t} terms at r = 5, meaning we do not need to multiply by extra factors of t due to resonance.

When the forcing is a sum of terms of the form P(t) e^{αt} where P(t) is a polynomial and α is not a root of the characteristic equation, we typically try a particular solution of the form Q(t) e^{αt}, where Q(t) has the same degree as P(t). Since the highest degree of P is 2, we should choose Q(t) to be a general quadratic:

Q(t) = A2 t^2 + A1 t + A0.

Thus a natural ansatz for the particular solution is yp(t) = (A2 t^2 + A1 t + A0) e^{5t}.

This form accommodates the constant term, the t-term, and the t^2-term in the forcing function, and, because there is no resonance with α = 5, no extra factors of t are required beyond the quadratic in t.

To connect to the given answer fragment, note that the expression (A2 t^2 + A1 t + A0) e^{5t} matches the general structure we derived for yp(t). If one were to evaluate constants A2, A1, and A0 by plugging into the differential equation, their values would be determined, but the question here specifically asks for the form of the particular solution, which is captured by this polynomial-coefficient times e^{5t} structure.

MAP2302 L15 section 4.5

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

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