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
Given: $u=e^{-\alpha^2 k^2 t} \sin (kx)$
and $u_x=k \times e^{-\alpha^2 k^2 t} \cos (kx)$
Now, take the second derivative of $u$ with respect to $x$.
$u_{xx}=-k^2 \times e^{-\alpha^2 k^2 t} \sin (kx)$ ...(1)
Now, take the second derivative of $u$ with respect to $t$.
$u_{t}=-\alpha^2 k^2 \times e^{-\alpha^2 k^2 t} \sin (kx)$
From equation (1), we have
$u_t=\alpha^2 u_{xx}$
This is the required relationship for the heat conduction.
Hence, it has been proved that the function $u=e^{-\alpha^2 k^2 t} \sin (kx)$ is a solution of the heat conduction equation $u_t=\alpha^2 u_{xx}$.
