Image processing with difference equation

I am trying to send an take an image and blur it using a difference equation, My code takes each individual pixel of the image, runs it through the difference equation and then stores the new pixel value into a new array that is displayed upon completion. I am just having trouble with completing the difference equation. The difference equation I have is

   (1-p)x[n] = y[n] - p*y[n-1]

I have this difference equation in my code for the most part, I am just running into a wall trying to set what x should be equal to. I was going along with the example from this webpage, second to last page. (https://www.csun.edu/~skatz/ece351/matlab_tutorial_2.pdf)

    function [  ] = Blur_( lion, p )%Lion is the image, p is the variable used in the difference equation, set this between .1-.9
  row = 2;    %//Starting point for Second While Loop Currently skips all the first element of each row, should change once run through diff equation.
          %//Setting row = 1 will grab each element of the array
  column = 1;   %//Starting point for first While Loop
  A = 'End of Row'; %//Display at end of each row

[x_1,y_1,~] = size(lion);% find dimensions of image

figure    %//Defines new figure to be displayed
colormap(gray(255));
image(lion) %// the image stored in gray is a columnXrow uint8 image
title('Original Image');%//Gives title to original image

tableA = zeros(y_1,x_1, 'uint8'); %create array of zeros for tableA to reduce overhead

while (p > 0 && p < 1)%//Test range of p from 0.1 to 1 incrementing by .1

      while (column < (y_1 + 1)) %// First while loop, used for the rows

          while (row < (x_1 + 1)) %//Second while loop, used to access each column of First while loops current row
                element = double(lion(column,row))%//This stores the current Array cell value in the variable element
                %--------THE PART THAT ISN'T WORKING-------%
                a = [1 -p]; %//this is for y[n] - p*y[n-1]
                b = [1-p 0]; %// for (1-p)x[n]     (will return and replace 1 with p once code is functioning)
                x = element*4;%This part I cant solve for. I am not sure how to calculate from difference equation I just used some arbitrary value with the current pixel as a place holder 
                new_pixel = filter(b,a,x); %difference equation is applied to the array
                tableA(column,row)= new_pixel; %//This creates the new array for the new image, storing new_pixel into current row/column
                %-------------------------------------------%
                row = row + 1; 
            end 

            row = 1; %Reset row to the start at the first element in the array on the next row
            disp(A); %//End of row in array
            column = column + 1; %increment column

        end %//end column while loop

        column = 1; %//Reset column to 1, this will allow us to test the next value of p for a new image.

        figure %//Defines a new figure to be displayed
        colormap(gray(255));
        image(tableA)%//displays the new image
        title('New Image');%//Gives the new image a title

        p = p + .2; %//Increment p until 1
  end %//end p while loop
  end

Answer 1

[Image Analyst]

I don't know what the code is doing. Why don't you simply blur with a single line of code in a call to conv2() or imfilter()? It would be so much easier.

[Steven Laburtz]

I am taking each individual Pixel and passing it through a difference equation, I am required to use this equation. I wish i could just pass it through a blurring function.

[Image Analyst]

but what is x and what is y in the equation? I'd guess y is the intensity of the input image, and n is the index of the pixel (like what column your pixel is on so it's like the "(x,y)" coordinate), but what is x in the equation? Is x the intensity of an output image? This is why I don't like it when people use single letter variable names - it's not very self-descriptive and we're left guessing what the variables represent.

[Steven Laburtz]

I replaced my variable names for the most part. For the difference equation that I gave. I was just given a transfer function which was

H(z)=(1-p)/(1-pz^(-1)) when (0 < p < 1)

So I just solved for the difference equation of this transfer function. y should be the output and x the input I believe, the n is the current pixel being passed through. in my code n == element

[Image Analyst]

I still don't quite understand it. Is pz really p*z? So pz^(-1) is really p/z? Whatever, sounds like you got it solved.

[Steven Laburtz]

It's all right, I will talk to my professor about this some more. Thanks for all the help you have provided!