Question

This problem was asked and closed before it is solved. I will ask it here again so it can hopefully be addressed
Solve
\[
\frac {dx}{y+z} =\frac {dy}{x+z}=\frac {dz}{x+y}
\]

Answer 1

@FibonacciChick666

Answer 2

hey @eliassaab do you have one of those computer answers you always give :)
I would like to see the answer

Answer 3

*

Answer 4

Hey I found something that might be useful to look at. I'm pretty sure it is above what I can comprehend.

Answer 5

http://math.stackexchange.com/questions/1096131/solve-this-pde-yzu-xzxu-yxyu-z-0

Answer 6

Like I was thinking a system would come in handy but I didn't know how to set it up... I also don't know how he got those equations he got. x'=y+z and so on...

Answer 7

I will post the solution after some responses.

Answer 8

@freckles, well for the solution above, the computer generated the pdf file from a latex file!!!

Answer 9

On of the solutions found above is graphed in the attached as the the intersections of two surfaces.

Answer 10

@ganeshie8

Answer 11

Cleverly done sir xD

Answer 12

Complete solution including the elimination of t.

Answer 13

wow, thanks guys! Turns out it was PDE. The teacher is just a jerk. I plan on using that solution to study though elias! Thank you everyone!

Answer 14

YW @FibonacciChick666

Answer 15

you want me to post the solution the teacher said?

Answer 16

yes. I would love to see it.

Answer 17

Note that \[\frac{dx}{z+y}=\frac{dy}{x+z}=\frac{dz}{x+y}=\frac{d(x-y)}{y-x}=\frac{d(x-z)}{z-x}=\frac{d(y-z)}{z-y}\] Hence we obtain solution via first integrals.\[ln|x-y|-ln|y-z|=C_1~~~~~~ln|x-z|-ln|y-z|=C_2\] This gives us two first integrals \[\Phi (x,y,z):=\frac{x-y}{y-z},~~~~~~~\Psi (x,y,z):=\frac{x-z}{y-z}.\] The general first integral is \[\Phi (x,y,z):=F(\frac{x-y}{y-z},\frac{x-z}{y-z}).\] Where F(u,v) is an arbitrary function. Indeed \(\Phi\) is the general solution of \[(z+y)\frac{\delta \Phi}{\delta x}+(x+z)\frac{\delta \Phi}{\delta y}+(x+y)\frac{\delta \Phi}{\delta z}=0.\] End

Answer 18

He's a jerk. And that word is not strong enough

Answer 19

No, he is not a jerk. Actually, I can deduce my solution from his method.

Answer 20

To put something on a test we have not even learned that is not in the spectrum of the class? Yes he's a jerk.

Answer 21

we never learned what a first integral even is

Answer 22

Then tell us you can ask anyone for help on this, encouraged us to ask for help even, from other teachers, graduate students, anyone, so long as they are not our classmates or him-yes, he is a jerk. then to teach above our heads and say halfway through lecture "Oh, I forgot you are not graduate students." not to mention giving less than 48 hours to complete 37 problems of this caliber. Jerk worthy.