Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
    
Can that be right? The denominator at the end is $0$.... –  Cameron Buie Aug 26 '12 at 18:54
    
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
show 9 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.