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