Solve: 2Z= (ax+y)²+b | Formation of PDE by eliminating arbitrary constants

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[@mamadetaslimtorabally7363]

This question was a bit complicated. However, isn't the last part q^3=px+qy because the RHS denominator is (q^2)^2.

[@rkchy]

the denominator is q² only

[@mamadetaslimtorabally7363]

@@rkchy OK sir. I understand. Thank you.

[@AbdurRahman-op8zn]

If q^2 is represented as 't'. Then instead of q^2 there should be t in the steps right ?

[@rkchy]

from where you are telling q² as t?

[@AbdurRahman-op8zn]

@rkchy dow^2 z / dow y^2 is represented as 't' right. I mean the notations p,q,r,s,t. Correct me if I'm wrong.

[@rkchy]

you may check from this: th-cam.com/video/HzvZJipNcNo/w-d-xo.htmlsi=ljalQMjIZf-gGCTi

[@suryaprakashsingh851]

Ok please correct me if I am wrong but isn't finding the value of b completely pointless
We can still get the partial differential equation
q^2 = px + qy using just the eq 2 and 3

[@rkchy]

ya maybe

[@KushSehgal18]

could you please explain how would we do that?

[@steve__vi]

@@KushSehgal18 take value of a from eq2 and put it in eq1

[@dudator9689]

@@steve__vi but b still stays there then

[@rashmirawat592]

Thank you so much

[@rkchy]

You're most welcome

[@reyazansari7175]

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