Solution for 16uxx=U₂0 0 subject to u(0,t)=u(z,t)=0,t>0 du u(x,0) = 0, 1 a) —sin (167) sin(x) 16 c) 1 b) = sin(16t) sin(x) 4 d) at It=0 4 sin (4t) sin(x) = sinx, 0
Solution for 16uxx=U₂0 0 subject to u(0,t)=u(z,t)=0,t>0 du u(x,0) = 0, 1 a) —sin (167) sin(x) 16 c) 1 b) = sin(16t) sin(x) 4 d) at It=0 4 sin (4t) sin(x) = sinx, 0
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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Transcribed Image Text:Solution for 16uxx=U₂0<x<,t> 0 subject to
u(0,t)=u(z,t)=0,t>0
du
u(x,0) = 0,
1
a) —sin (167) sin(x)
16
c)
1
b) = sin(16t) sin(x)
4
d)
at It=0
4
sin (4t) sin(x)
= sinx, 0<x< A
1
-sin (t) sin(x)
16
is ?.
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