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The solution of the heat equation Urr Ut, 0 < x < 1,t > 0, which satisfies the boundary conditions u (0, t) = 5 and u(1, t) = 1 and the initial condition u(x, 0) I has the form u(x, t) = v(x) +E bn sin(nrx)e n²t where n=1 v(x) = 5 – 4x and bn 10 f, (x – 1) sin(nn2) dx b) None of these O v(x) = 5 – 4x and br = 2 f, z sin(nax) dx 0(x) = 5 4x and bn = 10 fo (1 – 2) sin(nrx) da e v(x) = x + 5 and b, = 2 f, a sin(nn2) dx

The solution of the heat equation Urr Ut, 0 < x < 1,t > 0, which satisfies the boundary conditions u (0, t) = 5 and u(1, t) = 1 and the initial condition u(x, 0) I has the form u(x, t) = v(x) +E bn sin(nrx)e n²t where n=1 v(x) = 5 – 4x and bn 10 f, (x – 1) sin(nn2) dx b) None of these O v(x) = 5 – 4x and br = 2 f, z sin(nax) dx 0(x) = 5 4x and bn = 10 fo (1 – 2) sin(nrx) da e v(x) = x + 5 and b, = 2 f, a sin(nn2) dx

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 solution of the heat equation Urx Ut, 0 < x < 1,t > 0, which satisfies the boundary conditions u(0, t) = 5 and u(1, t) = 1 and the initial condition u(x, 0) I has the form u(x, t) = v(x)+E bn sin(nrx)et, where n=1 v(x) = 5 – 4 and bn = 10 f (x – 1) sin(nr2) dx b) None of these v(x) = 5 – 4x and b, = 2 f a sin(nnx) dx v(x) = 5 – 4x and br = 10 f, (1 – 2) sin(nr2) dr v(x) = x +5 and bn = 2 fo z sin(nn2) dx
Transcribed Image Text:The solution of the heat equation Urx Ut, 0 < x < 1,t > 0, which satisfies the boundary conditions u(0, t) = 5 and u(1, t) = 1 and the initial condition u(x, 0) I has the form u(x, t) = v(x)+E bn sin(nrx)et, where n=1 v(x) = 5 – 4 and bn = 10 f (x – 1) sin(nr2) dx b) None of these v(x) = 5 – 4x and b, = 2 f a sin(nnx) dx v(x) = 5 – 4x and br = 10 f, (1 – 2) sin(nr2) dr v(x) = x +5 and bn = 2 fo z sin(nn2) dx
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