Let us consider some molecules (or small particles, or microbes, etc) with concentration $c$. [Concentration is the number of particles per unit volume.] In the case that all particles are on a line, the concentration, $c=c(x,t)$, is the number of particles per unit length.
Now, consider that they are in motion with flux $J(x,t)$. [Flux is the number of particles passing across a certain point $x$ per unit time.]
Fickian diffusion dictates that
$$ J \sim \frac{\partial c}{\partial x} \Rightarrow J=-D\frac{\partial c}{\partial x}. \tag{1} $$
[The flux is proportional to the gradient of the concentration and particles flow from higher to lower concentration.]
Let us consider an interval from $x_1$ to $x_2$ and write the change in the number of particles
$$ \frac{\partial}{\partial t} \underbrace{\int_{x_1}^{x_2} c(x,t)\,dx}_\text{number of particles} = J(x_1,t) - J(x_2,t). $$
Let us now consider that the interval is infinitesimal $x_2-x_1 = \Delta x \to dx$ and get
$$ \frac{\partial}{\partial t} (c\,\Delta x) = J(x_1,t) - J(x_2,t) \Rightarrow \frac{\partial c}{\partial t} = \frac{J(x_1,t) - J(x_2,t)}{\Delta x} \rightarrow \frac{\partial c}{\partial t} = -\frac{\partial J}{\partial x}. $$
We may now use Fick’s law and obtain the Diffusion equation
$$ \frac{\partial c}{\partial t} =\frac{\partial}{\partial x} \left( D\frac{\partial c}{\partial x} \right). $$
In the case that $D$ is constant, we have the simpler equation
$$ \frac{\partial c}{\partial t} = D\,\frac{\partial^2 c}{\partial x^2}. \tag{2} $$
[This is a linear partial differential equation.]
See this tool for solving the diffusion equation online.
Exercise. (a) Show that the following is a solution of the diffusion equation (2),
$$ c(x,t) = \frac{Q}{2(\pi Dt)^{1/2}} e^{-x^2/(4Dt)},\quad t>0, \quad Q:\text{const.} $$