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.

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

类似问题

Determine the form of a particular solution for the given equation. Do not solve the equation. y″−24y′+144y=t2e12t+e12t+4sin(12t)

Can the method of undetermined coefficients together with superposition be applied to find a particular solution of the given equation? y″−3y′+5y=2cos(8t)−e3tt8+ 1 t

Determine the form of a particular solution to y″−10y′+26y=e2t+tsin5t−cos5t

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

Can the method of undetermined coefficients with superposition be used to solve the DE? y″−y′+y=(7e9t+t)5 a. no, since the coefficients of the DE are not constant b. no, because the right side of the equation is not the correct type of function c. no, because the DE is not linear d. yes

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

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单