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This question is about the heat equation. I did not understand where trying to solve cn. I did not understand why the coefficient is 5/2.

This question is about the heat equation. I did not understand where trying to solve cn. I did not understand why the coefficient is 5/2.

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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This question is about the heat equation. I did not understand where trying to solve cn. I did not understand why the coefficient is 5/2. 

u(x, 0) = 2 sin (Tx/2) – sin Tx + 4 sin 27x, 0< x < 2. Consider the conduction of heat in a rod 40 cm in length whose ends are maintained at 0°C for all t > 0. In each of Problems 9 through 12, find an expression for the temperature u(x, t) if the initial temperature distribution in the rod is the given function. Suppose that a? = 1. 9. и (х, 0) — 50, 0 < х < 40 > Answer Solution The heat conduction problem is formulated as = Ut, 0 < x < 40, t > 0; Uxx u(0, t) = 0, u(x, 0) = 50, u(40, t) = 0, t > 0; 0 < x < 40. %3| %3D sin nrx/40, with associated eigenvalues A, = n²7? /1600. The Assume a solution of the form u(x, t) = X(x)T(t). Following the procedure in this section, we obtain the eigenfunctions X, solutions of the temporal equations are Tn = e-^nt. Hence the general solution of the given problem is -n²n²t/1600 sin u(x,t) = Cne 40 n=1 The coefficients c, are the Fourier sine coefficients of u(x, 0) 50. That is, %3D 40 2 1 100 Cos nT dx f(x) sin L sin dx = 40 Cn 1...
Transcribed Image Text:u(x, 0) = 2 sin (Tx/2) – sin Tx + 4 sin 27x, 0< x < 2. Consider the conduction of heat in a rod 40 cm in length whose ends are maintained at 0°C for all t > 0. In each of Problems 9 through 12, find an expression for the temperature u(x, t) if the initial temperature distribution in the rod is the given function. Suppose that a? = 1. 9. и (х, 0) — 50, 0 < х < 40 > Answer Solution The heat conduction problem is formulated as = Ut, 0 < x < 40, t > 0; Uxx u(0, t) = 0, u(x, 0) = 50, u(40, t) = 0, t > 0; 0 < x < 40. %3| %3D sin nrx/40, with associated eigenvalues A, = n²7? /1600. The Assume a solution of the form u(x, t) = X(x)T(t). Following the procedure in this section, we obtain the eigenfunctions X, solutions of the temporal equations are Tn = e-^nt. Hence the general solution of the given problem is -n²n²t/1600 sin u(x,t) = Cne 40 n=1 The coefficients c, are the Fourier sine coefficients of u(x, 0) 50. That is, %3D 40 2 1 100 Cos nT dx f(x) sin L sin dx = 40 Cn 1...
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