## Calculus 8th Edition

The function $u=e^{-\alpha^2 k^2 t} \sin (kx)$ is a solution of the heat conduction equation $u_t=\alpha^2 u_{xx}$.
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 $u_t=\alpha^2 u_{xx}$.