Answer :
[tex]Log(az-1)=x+ay+b\\\\differentiating\ wrt\ x\\\\\frac{a}{az-1}\frac{dz}{dx}=1+a\frac{dy}{dx}\\\\\frac{az'}{az-1}=1+ay',\ \ \ \ ---(1)\\\frac{z'}{z-1/a}=1+ay'\\\\z'=z-1/a+(az-1)y'\\z'+(1-az)y'+1/a-z=0,\ \ \ --(2) [/tex]