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

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by jeninisha

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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*

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]

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