Question

form PDE by eliminating arbitrary function from lx+my+nz = f (x^2+y^2+z^2)

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Answer 1

Answer:

Form a PDE from the relation z = f1(ax + by) + f2(cx + dy). Formation of partial differential equations by elimination of arbitrary function φ from φ(u,v) = 0 where u and v are functions of x,y and z.

Answer 2

Given:

$lx+m y+n z=f\left(x^{2}+y^{2}+z^{2}\right)$

To find:

The partial differential equation by eliminating arbitrary function.

Solution:

$lx+m y+n z=f\left(x^{2}+y^{2}+z^{2}\right)$

$Differentiating with respect to \mathrm{x} and \mathrm{y} respectively, \begin{array}{l}l+n p=f^{\prime}\left(x^{2}+y^{2}+z^{2}\right)(2 x+2 z p) \\m+n q=f^{\prime}\left(x^{2}+y^{2}+z^{2}\right)(2 y+2 z q)\end{array} Dividing (2) by (3) \begin{aligned}\frac{l+n p}{m+n q} &=\frac{x+z p}{y+z q} \\(l+n p)(y+z q) &=(x+z p)(m+n q) . \\l y+n p y+l z q &=m x+m z p+n x q \\(m-n y) p+(n x-l z) q &=l y-m x .\end{aligned}$

This is the required partial differential equation.