Here is the basic equation for the gaussian equation or normal distribution.  x is the value of the variable and s is the standard deviation; μ is the mean and σ is the standard deviation.

(1/(σ*(Sqrt[2*Pi]))) 
      Exp[-((((((x - μ))^2)))/((2*σ)))]

((E^(-((x^2)/(2 s))))/((Sqrt[2 π]) s))

Here is a plot of the basic normal distrubtion with μ = 0 and σ=1.

Plot[(1/(1*(Sqrt[2*Pi]))) 
        Exp[-((((((x - 0))^2)))/((2)))], {x, -4, 
        4}]

Below is the difference of gaussians.  In both gaussians, μ = 0, however the σ's are different fro the two eqautions which allows the different not to sum to 0.  In the first case, the first gaussian will have a σ = 1/2 and the second gaussian will have a σ= 2.  Then the ratios will be changed gradually until the ratio is reversed.

Do[Plot[(1/((1*s)*(Sqrt[2*Pi]))) 
            Exp[-((((x^2)))/((2*(((1*s))^2))))] - (1/((1/
                      s)*(Sqrt[2*Pi]))) 
            Exp[-((((x^2)))/((2*(((1/s))^2))))], {x, -6, 
          6}], {s, 1/2, 2, 1/4}]

Below is a 2-dimensional gaussian function.

Plot3D[(1/((1)*(Sqrt[2*Pi]))) 
        Exp[-(((x^2) + (y^2))/((2*(((1))^2))))], {x, -6, 
        6}, {y, -6, 6}, PlotPoints -> 100, 
      PlotRange -> {0,  .4}]

Here is my attemt to make a 2  
        dimensional DOG.

Do[Plot3D[(1/((1*s)*(Sqrt[2*Pi]))) 
            Exp[-((((x^2) + 
                        (y^2)))/((2*(((1*s))^2))))] - (1/((1/
                      s)*(Sqrt[2*Pi]))) 
            Exp[-((((x^2) + 
                        (y^2)))/((2*(((1/s))^2))))], {x, -6, 
          6}, {y, -6, 6}, PlotPoints -> 75, 
        PlotRange -> {- .5,  .6}, 
        ViewPoint -> {1.539,  -2.840,  1.008}], {s, 1/2, 2, 
        1.5}]