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This is a question from the Sneddon book "Elements of PDE":

Prove that an equation of the "Clairaut" form $$ xu_x+yu_y+zu_z=f\big(u_x,u_y,u_z \big) \tag{1}$$ is always soluble by Jacobi's method.

First thing is that what I have read that the "Clairaut form " is $ z=xz_x+yz_y+f(z_x,z_y)$ . Then how will I proceed with it. Could anyone help me?

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Answer to :"HOW SHOULD YOU PROCEED?"
In $z=xz_x+yz_y+f(z_x,z_y)$,substitute $z_x=-p_1/p_3,z_y=-p_2/p_3$ so that you get:
$u_1x+u_2y+u_3z=u_3f(-u_1/u_3,-u_2/u_3)\implies u_1x+u_2y+u_3z=g(u_1,u_2,u_3) $
Now apply Jacobi's method:
$\frac{dx}{x-g_{u1}}=\frac{dy}{y-g_{u_2}}=\frac{dz}{z-g_{u_3}}=-du_1/u_1=-du_2/u_2=-du_3/u_3$
So it was how to proceed...

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