Why does this function satisfy $\frac{\partial p}{\partial t} - \frac{1}{2}\frac{\partial^2 p}{\partial z^2}=0 ?$

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I am working on a mock exam for Partial differential equations, and the exam claims that the following two functions

$p(z,t) = \frac{1}{2\lambda} e^{-\lambda z + \lambda^2 t / 2}$

and

$p(z,t) = \frac{1}{\sqrt{2\pi t}} e^{-z^2/(2t)}$

satisfy the equation

$\frac{\partial p}{\partial t} - \frac{1}{2}\frac{\partial^2 p}{\partial z^2}=0 ?$

But I don't think they do.

For the first one, my "solution" was to note that:

$\frac{\partial p}{\partial t} = -\frac{\lambda^2}{2}p(z,t)$ and $\frac{\partial^2 p}{\partial z^2} = \lambda^2p(z,t)$

Hence, the equation does not hold.

What am I missing?