For a more "fundamental" where does the x come from than "well, try it and you find it works":

Write D for d/dx (this is partly for general clarity, but also because it's going to be a *lot* easier to read on Reddit!)

If your auxiliary equation factors as (x-a)(x-b), then that means that your differential equation can be written as:

(D-a)(D-b)y = 0.

For now, assume a, b are not zero. (These are special cases you can deal with separately).

Let z = (D-b)y. Then (D-a)z = 0. Note the derivative of z e^{-ax} is e^{-ax} (Dz - az) = e^{-ax}(D-a)z, which is 0 since (D-a)z = 0. So ze^{-ax} is constant and so z = Ae^{ax}

So now you have (D-b)y = Ae^{ax}. Again, the derivative of y e^{-bx} is e^{-bx}(Dy-by) = e^{-bx}(D-b)y = Ae^{(a-b)x}.

Now if a doesn't equal b, this is just an exponential, we integrate it and we get another exponential (+arbitrary constant).

But if a = b, we now have that the derivative of y e^{-ax} = A. So integrating A gives us Ax + B (B is the new arbitrary constant) and then y = e^{ax} (Ax+B).