(b) Given that the boundary and initial conditions are as below: Boundary conditions: u(0,t) =u(1,t)=0, t>0 Initial conditions: u(x,0) = x(1– x), Osx<1. du(x,0) -= 0, 0 Tum (1 – (-1)ª] cos nnt sin nax n=1
(b) Given that the boundary and initial conditions are as below: Boundary conditions: u(0,t) =u(1,t)=0, t>0 Initial conditions: u(x,0) = x(1– x), Osx<1. du(x,0) -= 0, 0 Tum (1 – (-1)ª] cos nnt sin nax n=1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![(b)
Given that the boundary and initial conditions are as below:
Boundary conditions:
u(0,t) =u(1,t) =0, t>0
Initial conditions:
u(x, 0) = ;x(1 – x), O<x<1.
|
du(x,0)
= 0, 0<x< 1.
at
Show that the solution for the nontrivial case, which is for k = -22, is:
%3D
1
u(x, t) = 2un)3 (1 – (-1)"] cos nat sin nAx
n=1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F71a9673c-8a96-4e61-900b-4b5116b82ebc%2F0f358417-4ad6-4a04-8b1d-5b13d86867b6%2Faznya89_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(b)
Given that the boundary and initial conditions are as below:
Boundary conditions:
u(0,t) =u(1,t) =0, t>0
Initial conditions:
u(x, 0) = ;x(1 – x), O<x<1.
|
du(x,0)
= 0, 0<x< 1.
at
Show that the solution for the nontrivial case, which is for k = -22, is:
%3D
1
u(x, t) = 2un)3 (1 – (-1)"] cos nat sin nAx
n=1
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