Fin.
A fin is a solid body exposed to a fluid for the purpose of exchanging heat with the fluid. This discussion develops the differential equation for temperature in a fin that is thin and of uniform cross section.

Let $T(x)$ by the temperature along the fin. The fin exchanges heat with the fluid according to Newton's law of cooling

$$dq_{conv} = h Pdx [T(x) - T_{\infty}]$$

where

$dq_{conv}$	small amount of convection heat flow (W)
$h$	heat transfer coefficient (W/m$^2$/K)
$T_{\infty}$	fluid temperature (K)
$P$	perimeter of fin exposed to fluid (m)
$dx$	length of element of fin (m)

Consider a small element of the fin as shown in the figure. A steady energy balance for heat flow in this element is given by

$$q_x = q_{x+dx} + dq_{conv}$$

write as

$$- ( q_{x+dx} - q_x) - dq_{conv} =0$$

Let the heat flux leaving the element be given by one term of a Taylor series

$$q_{x+dx} = q_x + \frac{dq_x}{dx} dx$$

and let the heat flux at $x$ be described by Fourier's law:

$$q_x = - k A_c \frac{dT}{dx}.$$

Here $k$ is the conductivity (W/m/K) and $A_c$ (m$^2$) is the cross-section area of the fin. Now combine Fourier's law, the Taylor's series for $q_{x+dx}$, the energy balance, and Newton's law of cooling to obtain:

$$\frac{d}{dx} \left[ k A_c \frac{dT}{dx} \right] dx - h P dx [T(x) - T_{\infty}] = 0$$

Next if the conductivity, $k$, is constant, the cross-section area, $A_c$ is constant, divide by these. If the heat transfer coefficient $h$ is also contant along $x$, then divide by $dx$ to obtain:

$$\frac{d^2T}{dx^2} - \frac{hP}{kA_c} [T(x) - T_{\infty}] = 0$$

Finally, let $m^2 = (hP)/(kA_c)$ to give

$$\frac{d^2T}{dx^2} - m^2 [T(x) - T_{\infty}] = 0$$

This equation describes the heat transfer in a fin of uniform cross section.