Answer
$u_{1,x}=e^{-\alpha^2t}\cos x,\ u_{1,xx}=-e^{-\alpha^2t}\sin x$
$u_{1,t}=-\alpha^2e^{-\alpha^2t}\sin x$
$\alpha^2u_{1,xx}=u_{1,t}$
$u_{2,x}=\lambda e^{-\alpha^2\lambda^2t}\cos \lambda x,\ u_{2,xx}=-\lambda^2e^{-\alpha^2\lambda^2t}\sin \lambda x$
$u_{2,t}=-\alpha^2\lambda^2e^{-\alpha^2\lambda^2 t}\sin \lambda x$
$\alpha^2u_{2,xx}=u_{2,t}$