Change of variables in differential equation

[michalpp]

Homework Statement

I have to transform the following equation using variables (u,v,w(u,v))=(yz-x,xz-y, xy-z):

(xy+z)\frac{\partial z}{\partial x}+(1-y^2)\frac{\partial z}{\partial y}=x+yz.

Homework Equations

chain rule: <br /> \frac{dw}{dx} = \frac{\partial w}{\partial u} \frac{\partial du}{\partial dx} + \frac{\partial w}{\partial v} \frac{\partial dv}{\partial dx}<br />

The Attempt at a Solution

Using the chain rule and the product rule:

\frac{\partial w}{\partial x}=\frac{\partial z}{\partial x}(y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v})-\frac{\partial w}{\partial u}+z\frac{\partial w}{\partial v}

and a similar expression for \frac{\partial w}{\partial y}.
On the other hand

w=xy-z, so

\frac{\partial w}{\partial x}=y-\frac{\partial z}{\partial x}

(and similar for \frac{\partial w}{\partial y}), so

\frac{\partial z}{\partial x}(y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v})-\frac{\partial w}{\partial u}+z\frac{\partial w}{\partial v}=y-\frac{\partial z}{\partial x} and therefore:

\frac{\partial z}{\partial x}[y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1]=y+\frac{\partial w}{\partial u}-z\frac{\partial w}{\partial v} and

\frac{\partial z}{\partial y}[y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1]=x+\frac{\partial w}{\partial v}-z\frac{\partial w}{\partial u}.

After multiplying the given equation by

[y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1]

and writing

\frac{\partial z}{\partial x}[y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1] as y+\frac{\partial w}{\partial u}-z\frac{\partial w}{\partial v} I get

\frac{\partial w}{\partial v}(1-x^2-y^2-z^2-2xyz)=0.

If
[y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1]\neq 0

this equation is equivalent to the given one. But what if [y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1]=0?.

I'm kind of stuck at this point, cause in the first case

\frac{\partial w}{\partial v}=0 or (1-x^2-y^2-z^2-2xyz)=0.

I was thinking of a way to change (x,y,z) to (u,v,w) in the last equation, but without success. I also have no idea what to do in the second case. So what should I do now? Is this a correct method to solve a problem like this?

 

[michalpp]

I've found the answer to this problem in a book by B.P. Demidowich - it says
<br /> \frac{\partial w}{\partial v}=0<br />
So this means that I need a proof that
<br /> (1-x^2-y^2-z^2-2xyz)\neq 0<br />, or do I still miss something?

<br /> (1-x^2-y^2-z^2-2xyz)=(xy+z)z+(1-y^2)(-1)+x(x+yz)<br />
looks similar to
<br /> (xy+z)\frac{\partial z}{\partial x}+(1-y^2)\frac{\partial z}{\partial y}=x+yz.<br /> but I don't have other ideas.