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Z. The initial solution of the partial differential equation = 0 by separation of variable is a) z = X(x)+Y(y) b) z= X(x)– Y(y) z = X(x).Y(y) c) d) None of these

Z. The initial solution of the partial differential equation = 0 by separation of variable is a) z = X(x)+Y(y) b) z= X(x)– Y(y) z = X(x).Y(y) c) d) None of these

Advanced Engineering Mathematics
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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The initial solution of the partial differential equation by separation of variable is

az ôz ôx? Ox ôy Oz O by separation of The initial solution of the partial differential equation variable is a) z= X(x)+Y(y) b) z= X(x)– Y(y) c) Z = X(x).Y(y) d) None of these
Transcribed Image Text:az ôz ôx? Ox ôy Oz O by separation of The initial solution of the partial differential equation variable is a) z= X(x)+Y(y) b) z= X(x)– Y(y) c) Z = X(x).Y(y) d) None of these
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The solution of one dimensional heat equation

7. The solution of one dimensional heat equation ди = 0,u(x,0) = 5sin x – 2 sin 27X at t-0 u(0,1) %3D 0, и(3,1) %3D 0, a) u(x,t) = 5 sin(zx).cos T ct –2sin(2x).cos 2n ct b) u(x,t)= 5sin(x).cos T Ct +2 sin(x).coS T ct c) u(x,t)=5sin(x).cos T ct – 2 sin(27x).cos 27 ct d) None of these
Transcribed Image Text:7. The solution of one dimensional heat equation ди = 0,u(x,0) = 5sin x – 2 sin 27X at t-0 u(0,1) %3D 0, и(3,1) %3D 0, a) u(x,t) = 5 sin(zx).cos T ct –2sin(2x).cos 2n ct b) u(x,t)= 5sin(x).cos T Ct +2 sin(x).coS T ct c) u(x,t)=5sin(x).cos T ct – 2 sin(27x).cos 27 ct d) None of these
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