Saying that the solution to the homogeneous equation is of the form at+ b means, since G(t, z) must satisfy the equation for all t except possibly t= z, that G(t, z) is of the form At+ B for t< z and Ct+ D where t> z. At t= z, of course, the two must match so you must have G(z, z)= Az+ B= Cz+ D. G(0, z)= A(0)+ B= B= 0 (0< z, of course). G(1, z)= C(1)+ D= C+ D= 1, D= 1- C. Putting those into Az+ B= Cz+ D, we have Az= Cz+ 1- C= C(z- 1)+ 1.

You will need one more condition to find C and so A and D: the difference between the two derivatives, taken from the left and right, at t= z, must be 1:

for t< z, dG/dt= A and for t> z, dG/dt= C: A- C= 1 so that A= C+ 1. Putting that int Az= C(z- 1)+ 1 gives (C+1)z= C(z- 1)+ 1 so Cz+ z= Cz- C+ 1. The "Cz" terms cancel: z= 1- C so C= 1-z. then A= C+ 1= 1- z+ 1= -z, D= 1- C= 1- (1-z)= z, and we already had B= 0: