A question on Lagrange's Method for solving partial differential equation

Question

The question is to solve $(y-z)p+(z-x)q=(x-y)$ where $p=\frac{\partial z}{\partial x}$ and $q=\frac{\partial z}{\partial y}$

The solution I am referring to has this following line:

$\frac{dx}{y-z}=\frac{dy}{z-x}=\frac{dz}{x-y}=\frac{dx+dy+dz}{(y-z)+(z-x)+(x-y)}$

Though I am perfectly fine till the last but one equality but how did we go about doing the last bit, I am not sure. Plus this looks so very strange that I am not able to understand much.I am coming across at many more place still stranger stuff like $\frac{\sum xdx}{\sum x(y-z)}$

Help appreciated Soham

I am teaching myself Pde and as for all practical matters I am a noob — Soham, Aug 26 '12 at 18:43
Consider simply $k=\frac ab=\frac cd=\frac {a+c}{b+d}$. This is true simply because $a=kb\ $ and $c=kd\ $ so that the sum $a+c=kb+kd$. — Raymond Manzoni, Aug 26 '12 at 18:52
Oops I didn't notice that (they probably conclude that $dx+dy+dz$ must be zero too in this case !) — Raymond Manzoni, Aug 26 '12 at 18:56
@RaymondManzoni That exactly was my problem... at the first glance, what you said looks okay,the same also I thought about, but the denom at the end is $0$... — Soham, Aug 26 '12 at 18:56
Apparentlly after some googling I found it-its called method of multipliers...Its strange and I would love to see the motivation and how things developed like this. — Soham, Aug 26 '12 at 18:58
do they conclude that $dx+dy+dz=0$ so that $x+y+z=$constant ? — Raymond Manzoni, Aug 26 '12 at 18:59
I must admit that it's not very academic ;-) but I understand the idea of the 'needed undeterminate form' $0\over 0$. Of course you could do it the way I did : conclude the same way that $dx+dy+dz=k(y-z+z-x+x-y)=0$ and $x+y+z=$constant without needing the indetermination. — Raymond Manzoni, Aug 26 '12 at 19:05
@RaymondManzoni And can we also say the same thing about $(xdx +y dy+ zdz)/(...) $ ? — Soham, Aug 26 '12 at 19:07
the denominator would be $x(y-z)+y(z-x)+z(x-y)=0$ too so that the same trick is at work, yes. — Raymond Manzoni, Aug 26 '12 at 19:10
I should add that my $k$ should rather be written $dt$ (it must be a differential operator and $t$ represents usually the 'parametrization' of the curve in the method of characteristics). — Raymond Manzoni, Aug 26 '12 at 19:47

A question on Lagrange's Method for solving partial differential equation

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