# Re: proving DE not separable (partials technique???)

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Posted by jonathan burros on September 11, 2002 at 13:01:01:

In Reply to: proving DE not separable (partials technique???) posted by Colin on September 11, 2002 at 11:18:06:

: A while back (see 8/8 posting) Subhotosh made a statement about a partials technique used to prove the unseparability of a DE. Can anyone explain this?

: I have a problem:
: dy/dx = (y - 4x)/(x - y)

: I need to prove it is not separable. How can I do it formally? It is pretty clear if I try it, but I don't know if that is a sufficient "proof" to show a couple failed attempts...

: Colin

You can separate it by substituting

y=ux where u is a differential function of x
to be determined

dy/dx = u + x du/dx

(y - 4x)/(x - y)

= (u -4)/(1-u)

Substitute into D.E.

u + x du/dx = (u -4)/(1-u)

Then x du/dx = (u^2 -4)/(1-u)

You can now separate the variables

[(1-u)/(u^2 - 4)]du = dx/x

Now express the left side in partial fractions

And you can integrate both sides

I leave this to you

Note:When you use partial functions

you get

dx/x = (-3/4)[du/(u+2)] -(1/4)[du/(u-2)]

Integrate both sides

ln|x| = (-3/4)ln|u+2| -(1/4)ln|u-2| -(1/4)ln|C|

4ln|x| = -[3ln|u+2|+ln|u-2|+ln|C|]

ln|x^4|=ln[1/|C(u+2)^3(u-2)|]

Take antilogarithms

|x^4|= 1/|C(u+2)^3(u-2)|

|x^4 C(u+2)^3(u-2)| = 1

+-C x^4(u+2)^3 (u-2) = 1

c x^3 (u+2)^3 x(u-2) = 1

c (ux + 2x)^3 (ux - 2x) = 1

Substitute y = ux

c(y + 2x)^3 (y - 2x) = 1

Note: C and c are arbitrary constants

c = +-C

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