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## Homework Statement

The heat equation for an

**infinitely**long rod is shown as:

[tex]

\alpha^2 \frac{\partial^2}{\partial x^2}u(x,t) = \frac{\partial}{\partial t}u(x,t)

[/tex]

[tex]

u(0,t) = u(L,t) = 0,\ \forall \ t > 0

[/tex]

[tex]

u(x,0) = sin(\pi x) \ \forall \ 1 < x < 2

[/tex]

[tex]

u(x,0) = 0\ \forall \ otherwise

[/tex]

[tex]

\alpha^2 = 0.1

[/tex]

**2. The attempt at a solution**

I know that the solution of this problem is a Fourier sine series:

u(x,t)= sum (n=0 to infinity) B_n * sin ((n pi x)/L)

However, I am having problem trying to determine the coefficient Bn:

[tex]

B_n = \frac{2}{L}\int_{0}^{L} sin(\pi x) sin(\frac{n \pi x}{L}) dx

[/tex]

Since the function u(x,0)= sin(pi x) for 10<x<11 I'm not sure if I should aproach this as:

[tex]

B_n = \frac{2}{L}\int_{1}^{2} sin(\pi x) sin(\frac{n \pi x}{L}) dx

[/tex]

where L is a large number

OR

[tex]

B_n = \frac{2}{2}\int_{1}^{2} sin(\pi x) sin(\frac{n \pi x}{2}) dx

[/tex]

Do I go ahead with the second one and ignore that the rod is infinitely long since u(x,0) = 0 for all other values or is this a mistake?

I apologise that some of the equations of the post look like this but I couldn't get my tex brackets to work.

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