Consider the initial boundary value problem (IBVP) for the wave equation: a²u a²u 4 at2 0 0 (1) и(0, t) —D и(п, t) %3D 0, t > 0 (2) u(x,0) = 0,u¿(x, 0) = sin x ,0 < x < n (3) The series u(x, t) = E-[An cos(2nt) + B, sin(2nt)] sin(nx) satisfies the wave equation (1) and the boundary conditions (2). By using the initial conditions given in (3) determine the coefficients A, and B,, and then decide which of the following is the solution of the IBVP (1) – (3). A u(x, t) = sin 4t sin 2x В u(x,t) = cos 2t sin 2x COs u(x,t) = sin t sin x D u(x,t) = cos 2t sin x E u(x, t) = sin 2t sin x

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.6: Additional Trigonometric Graphs
Problem 78E
Question
Consider the initial boundary value problem (IBVP) for the wave equation:
a²u
a²u
4
at2
0 <x < T,
t > 0 (1)
и(0, t) —D и(п, t) %3D 0,
t > 0
(2)
u(x,0) = 0,u¿(x, 0) = sin x ,0 < x < n (3)
The series u(x, t) = E-[An cos(2nt) + B, sin(2nt)] sin(nx) satisfies the wave
equation (1) and the boundary conditions (2). By using the initial conditions given in (3)
determine the coefficients A, and B,, and then decide which of the following is the
solution of the IBVP (1) – (3).
A
u(x, t)
= sin 4t sin 2x
В
u(x,t) = cos 2t sin 2x
COs
u(x,t) = sin t sin x
D
u(x,t) = cos 2t sin x
E
u(x, t) = sin 2t sin x
Transcribed Image Text:Consider the initial boundary value problem (IBVP) for the wave equation: a²u a²u 4 at2 0 <x < T, t > 0 (1) и(0, t) —D и(п, t) %3D 0, t > 0 (2) u(x,0) = 0,u¿(x, 0) = sin x ,0 < x < n (3) The series u(x, t) = E-[An cos(2nt) + B, sin(2nt)] sin(nx) satisfies the wave equation (1) and the boundary conditions (2). By using the initial conditions given in (3) determine the coefficients A, and B,, and then decide which of the following is the solution of the IBVP (1) – (3). A u(x, t) = sin 4t sin 2x В u(x,t) = cos 2t sin 2x COs u(x,t) = sin t sin x D u(x,t) = cos 2t sin x E u(x, t) = sin 2t sin x
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