I want to show that $u(x, t) = e^{-\alpha^2k^2t}\sin(kx)$ is a solution for $u_t = \alpha^2u_{xx}$.
I did the following:
$$u_x = e^{-\alpha^2k^2t}\cos(kx)k$$
$$u_{xx} = -e^{-\alpha^2k^2t}\sin(kx)k^2$$
$$u_{t} = -\alpha^2k^2e^{-\alpha^2k^2t}\sin(kx)$$
so, plugging $u_{xx}$ into the equation $u_t = \alpha^2u_{xx}$ gives the following:
$$u_t = \alpha^2[-e^{-\alpha^2k^2t}\sin(kx)k^2]$$
after rearranging, we get:
$$u_{t} = -\alpha^2k^2e^{-\alpha^2k^2t}\sin(kx)$$
Is this all I have to do?