# Not by separating variables

• Nov 21st 2009, 03:56 AM
phycdude
Not by separating variables
Here's the equation $\frac{dx}{\frac{ay}{x^2+y^2}}=\frac{dy}{\frac{ax}{ x^2+y^2}}$
how is this one done separating variables seems impossible
• Nov 21st 2009, 06:17 AM
Jester
Quote:

Originally Posted by phycdude
Here's the equation $\frac{dx}{\frac{ay}{x^2+y^2}}=\frac{dy}{\frac{ax}{ x^2+y^2}}$
how is this one done separating variables seems impossible

Doesn't $x^2+y^2$ cancel in each term? Was this the original problem?
• Nov 21st 2009, 10:38 AM
phycdude
Quote:

Originally Posted by Danny
Doesn't $x^2+y^2$ cancel in each term? Was this the original problem?

well thats what beats me too , i have no idea whether that would be legitimate ,
now this was not the question itself, i was just wondering, but here is what drove me to ask ,
MathBin.net - Untitled
if u like i can post the stuff here, im stumped.
• Nov 22nd 2009, 04:46 AM
HallsofIvy
Quote:

Originally Posted by phycdude
well thats what beats me too , i have no idea whether that would be legitimate ,
now this was not the question itself, i was just wondering, but here is what drove me to ask ,
MathBin.net - Untitled
if u like i can post the stuff here, im stumped.

Yes, of course that's legitmate. Your equation is simply [tex]\frac{dx}{y}= \frac{dy}{x}[quote] which is the same as xdx= ydy.
• Nov 22nd 2009, 06:39 AM
phycdude
[QUOTE=HallsofIvy;409885]Yes, of course that's legitmate. Your equation is simply [tex]\frac{dx}{y}= \frac{dy}{x}
Quote:

which is the same as xdx= ydy.
thanks, well i was wondering since pathlines are given by $\frac{dx}{dt} = u$ and the same for v and w with dx replaced by dy and dz respectively , how would i integrate to obtain the trigonometric solutions here mathbin.net/37252