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Whoa! How did they do this ?

[SOLVED] Whoa! How did they do this ?

Hi,

I am trying to teach myself partial diff equations and I'm stuck with this small step which I do not understand. I hope someone will help me.

Homework Statement
Solve the Partial Differential Equation
[tex](x^2-y^2-z^2)p+2xyq=2xz[/tex]


Solution given in the book

Subsidiary equations are

[tex]\frac{dx}{x^2-y^2-z^2}=\frac{dy}{2xy}=\frac{dz}{2xz}[/tex]

The step which I do not understand is this :

Using multipliers x, y, z. we get:
each fraction=[tex]\frac{xdx+ydy+zdz}{x(x^2+y^2+z^2)}[/tex]

Rest of the solution I understand but I don't get that how they did this particular step. How did they get this fraction ?

Thanks for your time and effort:smile:
 

Answers and Replies

tiny-tim
Science Advisor
Homework Helper
25,789
249
… ratios …

The step which I do not understand is this :

Using multipliers x, y, z. we get:
each fraction=[tex]\frac{xdx+ydy+zdz}{x(x^2+y^2+z^2)}[/tex]

Rest of the solution I understand but I don't get that how they did this particular step. How did they get this fraction ?
Hi Google_Spider! :smile:

It's easy algebra - but it looks extremely unlikely the first time you see it! :eek:

If the ratio X/A = Y/B = a, then you can add any multiples together on top and bottom, and the ratio is the same:

(pX + qY)/(pA + qB) = (paA + qaB)/(pA + qB) = a(pA + qB)/(pA + qB) = a.​

And the same works for X/A = Y/B = Z/C = a, and so on.

In this case, put p = x, q = y, r = z, and see what happens … :smile:
 
Thanks tiny-tim!:smile:
 

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