The diffusion equation is a partial differential equation. The unknown quantity is a function $C(x,t)$. To complete the problem statement you need to specify an initial condition (at $t=0$) and boundary conditions. I'm guessing that your boundary conditions are at infinity, so we take
$$ C(x,t) \rightarrow 0,\ x\rightarrow \pm \infty. $$
We take a delta function initial condition:
$$ C(x, 0) = \delta(x). $$
The equation can be solved by using the Fourier transform:
$$ C(x, t) = \int_{-\infty}^{\infty} \frac{\mathrm{d}k}{2\pi}\ \mathrm{e}^{i k x} C_k(t). $$
The inverse transform is
$$ C_k(t) = \int_{-\infty}^{\infty} \mathrm{d}x\ \mathrm{e}^{-i k x} C(x, t). $$
So the transform of the initial condition is
$$ C_k(0) = 1. $$
Substituting $C(x,t)$ in the diffusion equation gives
$$ \int_{-\infty}^{\infty} \frac{\mathrm{d}k}{2\pi}\ \mathrm{e}^{i k x} \left(
\dot{C}_k(t) + D k^2 C_k(t) \right) = 0.$$
This simplifies to
$$ \dot{C}_k(t) + D k^2 C_k(t) = 0, $$
with the solution
$$ C_k(t) = C_k(0) \mathrm{e}^{- D k^2 t} = \mathrm{e}^{- D k^2 t}. $$
Putting it all together
$$ C(x, t) = \int_{-\infty}^{\infty} \frac{\mathrm{d}k}{2\pi}\ \mathrm{e}^{i k x} \mathrm{e}^{- D k^2 t}, $$
and all that's left is to do the $k$ integral. Note that the $k$ integral is a Gaussian so, with a little massaging, you can do it with the formula
$$ \int_{-\infty}^\infty \mathrm{d}y\ \mathrm{e}^{-y^2} = \sqrt{\pi}. $$
You should get
$$ C(x,t) = \frac{1}{\sqrt{4 \pi D t}} \mathrm{e}^{-\frac{x^2}{4 D t}}. $$