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August 20th, 2017, 12:21 AM   #1
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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.
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August 25th, 2017, 04:49 AM   #2
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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.
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October 15th, 2017, 02:03 AM   #3
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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
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October 16th, 2017, 07:35 PM   #4
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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 $(dy)^2 = (dy)\cdot (dy)$.
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October 17th, 2017, 02:42 AM   #5
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That clarified everything. Thank you so much.
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