Question

form the partial differential equation from log(az-1)=x+ay+b where a and b are arbitary constants

Answer 1

log  (az - 1 )  = x + a y + b
  a z - 1  = exp (x+ ay + b)
taking partial differentials
  a  dz  =  exp (x + a y + b)  .  [dx  + a dy]

 a  dz   =  exp(x+ay+b) [ dx  +  a  dy  ]
             =  (az -1)  [ 1 +a  dy/dx]
   a  dz / (a z - 1)    =  1 dx + a dy
or    a/(az-1)   dz/dx    = 1  + a dy/dx

Answer 2

log(az-1)=x+ay+b.
 Differentiating partially with respect to x and y we get
1 / az-1 * ap = 1 ——->(1)
1 / az-1 * ap = a ——->(2)
Dividing (2) by (1) we get
 q/p = a
 From equation (1) ap = az-1
 a(z-p) = 1
q(z-p) =p
Answer :q(z-p)=p