this Q
Just substitute the function into the heat equation..
$\displaystyle \frac{\partial u}{\partial t}=-\alpha^2k^2\exp\left(-\alpha^2k^2t\right)\sin\left(kx\right)$
$\displaystyle \frac{\partial^2 u}{\partial x^2}=-k^2\exp\left(-\alpha^2k^2t\right)\sin\left(kt\right)$
It should be evident now that $\displaystyle u\left(x,t\right)=\exp\left(-\alpha^2k^2t\right)\sin\left(kx\right)$ satisfies the heat equation.
Does this help?
$\displaystyle u=e^{-\alpha^2k^2t}\sin(kx)$
$\displaystyle u_t=-\alpha^2k^2e^{-\alpha^2k^2t}\sin(kx)$
$\displaystyle u_x=e^{-\alpha^2k^2t}\cos(kx)\cdot k$
$\displaystyle u_{xx}=-e^{-\alpha^2k^2t}\sin(kx)\cdot k^2$
$\displaystyle \alpha^2u_{xx}=\alpha^2(-e^{-\alpha^2k^2t}\sin(kx)\cdot k^2)$
$\displaystyle = -\alpha^2k^2e^{-\alpha^2k^2t}\sin(kx)=u_t\quad \blacksquare$