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
We are given that $u=e^{-\alpha^2 k^2 t} \sin (kx)$
and $u_x=(k) e^{-\alpha^2 k^2 t} \cos (kx)$
Second derivative of $u$ with respect to $x$ gives:
$u_{xx}=-k^2 \times e^{-\alpha^2 k^2 t} \times \sin (kx)$ ...(A)
Also, the second derivative of $u$ with respect to $t$ gives:
$u_{t}=-\alpha^2 k^2 \times e^{-\alpha^2 k^2 t} \times \sin (kx)$
We can see from the equation (A) that
$u_t=\alpha^2 u_{xx}$
Hence, it has been verified that the function $u=e^{-\alpha^2 k^2 t} \sin (kx)$ is a solution of the heat conduction equation $u_t=\alpha^2 u_{xx}$.
