Not by separating variables

phycdude · #1 $Old$ November 21st, 2009, 04:56 AM

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

Danny · #2 $Old$ November 21st, 2009, 07:17 AM

Quote:

Originally Posted by phycdude $View Post$

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?

phycdude · #3 $Old$ November 21st, 2009, 11:38 AM

Quote:

Originally Posted by Danny $View Post$

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.

HallsofIvy · #4 $Old$ November 22nd, 2009, 05:46 AM

Quote:

Originally Posted by phycdude $View Post$

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 [math]\frac{dx}{y}= \frac{dy}{x}[quote] which is the same as xdx= ydy.

phycdude · #5 $Old$ November 22nd, 2009, 07:39 AM

[quote=HallsofIvy;409885]Yes, of course that's legitmate. Your equation is simply [math]\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