Question

Form the PDE of the following by eliminating arbitrary functions

phi(z^2-xy,x/z)

Answer 1

$\Phi(z^2-xy,x/z)=0$ (1)

We set $u=z^2-xy$; $v=\frac xz$.

Then the operator d on (1) leads to:

$\displaystyle d\Phi(u,v)=\frac{\partial \Phi(u,v)}{\partial u}du+\frac{\partial \Phi(u,v)}{\partial v}dv$=0.

Thus

$0=du=d(z^2-xy)$ (3)

$0=dv=d(\frac{x}{z})$ (4)

From (3) and (4) we have:

$2zdz-ydx-xdy=0$

$\displaystyle \frac1zdx-\frac{x}{z^2}dz=0$

Thus getting rid of dx we obtain:

$\displaystyle dx=\frac xzdz$

$\displaystyle (2z-y\frac xz)dz=xdy$

$\displaystyle dz=\frac {dy}{\frac{2z^2-yx}{xz}}=\frac{dx}{\frac{x^2}{xz}}$

$\displaystyle \frac{dz}{xz}=\frac {dy}{2z^2-yx}=\frac{dx}{x^2}$

Finally we obtain the desired PDE for $z(x,y)$:

$\displaystyle x^2\frac{\partial z}{\partial x}+(2z^2-xy)\frac{\partial z}{\partial y}=xz$

Answer: $\displaystyle x^2\frac{\partial z}{\partial x}+(2z^2-xy)\frac{\partial z}{\partial y}=xz$.