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August 19th, 2017, 11:21 PM  #1 
Newbie Joined: Apr 2017 From: India Posts: 17 Thanks: 0  Order and degree of a differential equation
I am unable to find the order and degree of the following differential equation: $x^2(dx)^2 + 2xy dx dy + y^2(dy)^2  z^2(dz)^2 = 0$ My approach: Take the term $z^2(dz)^2$ to right hand side and then divide throughout by $(dz)^2$ we get $$\mathrm{\dfrac{x^2(dx)^2}{(dz)^2} + \dfrac{2xy\ dx \ dy}{(dz)^2} +\dfrac{y^2(dy)^2}{(dz)^2} = z^2}$$ By seeing this, we can deduce that order is 1 and degree 2. But when I am looking at the middle term it contains dx dy/(dz)^2 term which is a term of order 2 and hence degree 1. Well am I right here? Is there any other way of finding the degree and order of this total differential equation? Do suggest, if any. 
August 25th, 2017, 03:49 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,647 Thanks: 680 
Since "dx", "dy", and "dz" are differentials, you shouldn't interpret "$\displaystyle (dx)^2$", "$\displaystyle (dy)^2$", and "(dz)^2" as powers  those indicate second derivatives. This equation is of second order and first degree.
Last edited by skipjack; August 25th, 2017 at 06:49 AM. 

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