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{\bf Question}

A thin wire is stretched along the x-axis and is heated by an
electric current.  The temperature T of the wire varies with x and
satisfies the differential equation:

$$\frac{d^2T}{dx^2}-p^2T = -k,$$

where p and k are nonzero constants.  Put $T=\frac{k}{p^2}+y$ and
show that y satisfies the differential equation:

$$\frac{d^2y}{dx^2}-p^2y = 0.$$

Hence show that the general solution of the equation for T is:

$$T = \frac{k}{p^2}+Ae^{px}+Be^{-px},$$

where A and B are constants.  Find the values of A nad B given
that Y satisfies the conditions $T = 0$ at $x = a$ and also $T =
0$ at $x = -a$ (where $a \not= 0$), and show that in this case

$$T = \frac{k}{p^2} \left ( 1 - \frac{\cosh px}{\cosh pa}\right)$$



{\bf Answer}

$\frac{d^2T}{dx^2} -p^2T = -k$ \hspace{1in} (*)

 If \hspace{.25in} $T = \frac{k}{p^2} + y,$ then (since $\frac{k}{p^2}$  is a
constant)

$\rightarrow \frac{dT}{dx} = \frac{dy}{dx}$   and
 $\frac{d^2T}{dx^2} = {d^y}{dx^2}$

Substituting into  (*)  gives \hspace{.25in} $\frac{d^2y}{dx^2}
-p^2 \left( \frac{k}{p^2}+y \right)  =  -k $

and so \hspace{.25in} $\frac{d^2y}{dx^2} - p^2y = 0 $ \hspace{1in}
(**)

auxiliary equation for (**) \hspace{.25in} $\lambda^2 -p^2  =  0 $

which has two distinct real roots: \hspace{.1in} $\lambda_1,
\lambda_2 =  \pm p$

General solution for (**): \hspace{.1in} $y = Ae^{px} + Be^{-px}$

Hence the general solution for (*) is given by

$T = \frac{k}{p^2} + y = \frac{k}{p^2} + Ae^{px} + Be^{-px}$

Boundary conditions:

$T(a) = 0 $ gives $ = \frac{k}{p^2} + Ae^{pa} + Be^{-pa}$ (1)

$T(-a) = 0 $ gives $ = \frac{k}{p^2} + Ae^{-pa} + Be^{pa}$ (2)

Equation (1) - Equation(2) $A(e^{pa} - e^{-pa)) +
B(e^{-pa}-e^{pa}} = 0$

and so $A(e^{pa}-e^{-pa})=B(e^{pa}-e^{-pa})$

from which we deduce that $A = B$ (note $pa \not= 0$ by
assumption, and so $(e^{pa}-e^{-pa}) \not= 0$)

Therefore $T =  \frac{k}{p^2} + y = \frac{k}{p^2} + A(e^{px} +
e^{-px}) = \frac{k}{p^2} + 2A \cosh (px)$

Boundary condition: $T(a) = 0$ gives

$0 = \frac{k}{p^2} + 2A \cosh (pa) \Rightarrow A = \frac{A}{2p^2
\cosh(pa)}$

Hence the particular solution: $$T = \frac{k}{p^2} \left( 1 -
\frac{ \cosh(px)}{ \cosh(pa)} \right)$$



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