
Differential Equations Ordinary and Partial Differential Equations Math Forum 
 LinkBack  Thread Tools  Display Modes 
August 20th, 2017, 12:21 AM  #1 
Member Joined: Apr 2017 From: India Posts: 34 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, 04:49 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
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 07:49 AM. 
October 15th, 2017, 02:03 AM  #3 
Member Joined: Apr 2017 From: India Posts: 34 Thanks: 0 
The source I am referring to says it is differential equation with order 1 and degree 2. Well I think there is some sort of confusion. Could someone please elaborate that for me?. I am totally confused 
October 16th, 2017, 07:35 PM  #4 
Senior Member Joined: Sep 2016 From: USA Posts: 502 Thanks: 280 Math Focus: Dynamical systems, analytic function theory, numerics 
Indeed this is order 1 and degree 2. It is order 1 because it has 3 differential terms: $dx,dy,dz$ and each is a first order differential. It is degree 2 because the highest degree that any of these differentials appears is to the 2nd power (in this case all 3 are raised to the 2nd power). Country boy's post would be correct if $(dy)^2$ means $d(dy)$ but it doesn't. It means literally the square of the differential $(dy)^2 = (dy)\cdot (dy)$. 
October 17th, 2017, 02:42 AM  #5 
Member Joined: Apr 2017 From: India Posts: 34 Thanks: 0 
That clarified everything. Thank you so much.


Tags 
degree, differential, differential equation, equation, order 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Degree of differential equation  panky  Calculus  1  April 26th, 2017 05:14 AM 
First order differential equation  BonaviaFx  Differential Equations  8  March 29th, 2015 09:20 AM 
Differential Equation of 4th degree w/o initial values  Thelair  Differential Equations  1  November 19th, 2013 08:24 AM 
RK4 method applied to a second degree differential equation  engineeringstudent  Differential Equations  0  March 21st, 2012 02:08 AM 
RK4 method applied to a second degree differential equation  engineeringstudent  Differential Equations  0  December 31st, 1969 04:00 PM 