Question

The partial differential equation obtained from z=a2x+ay2+bz=a2x+ay2+b by eliminating the arbitrary constants is 

Select one:

a. 4y2p−q2=04y2p−q2=0

b. p−4q=0p−4q=0 

c. yp−xq=0yp−xq=0

d. None of these 

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Answer 1

Answer:

I guess answer is none of these

Answer 2

The partial differential equation is Z=$a^{2}x+ay^{2} +bz=a^{2} x+ay^{2} +b$

Let partial differentiation of z with respect to x is $a^{2} +2ay+b$

partial differentiation with respect to y is$a^{2} +2ay$

$p=a^{2}$

$q=2ay$

sub a=$\frac{q}{2y}$ in p=$a^{2}$

$p= (\frac{q}{2y} )^{2}$$p=\frac{q^{2} }{4y^{2} }$

$4y^{2} p=q^{2}$

$4y^{2} p-q^{2} =0$

Therefore ,the option a is correct.

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