Answer
The function $u=e^{-\alpha^2 k^2 t} \sin (kx)$ is a solution of the heat conduction equation $u_t=\alpha^2 u_{xx}$.
Work Step by Step
Here, we have $u_x=k e^{-\alpha^2 k^2 t} \cos (kx)$
We need to take the second derivative of $u$ with respect to $x$.
$u_{xx}=-k^2 \times e^{-\alpha^2 k^2 t} \sin (kx)$ ...(a)
and
$u_{t}=-\alpha^2 k^2 \times e^{-\alpha^2 k^2 t} \sin (kx)$...(b)
From the above two equations (a) and (b), we have
$u_t=\alpha^2 u_{xx}$
This is the required relationship for the heat conduction.