A two Variable PDE

I'd like to know the answer myself. Does [itex]\frac{xU_x}{U}-\frac{xV_x}{V}=\frac{yU_y}{U}-\frac{yV_y}{V}[/itex] make it any easier? I've only dabbled in PDEs myself, but that seems like a separation of variables problem, when rephrased like that.

EDIT: Mathematica gives an error saying the answer is indeterminate, because there are more dependent variables than equations.

 

So you have:
x d/dx ln (U) -y d/dy ln(U) -1 = xd/dx ln (V) -y d/dy ln(V) +1
x d/dx (ln(U/V))=y d/dy (ln(U/V))+2

Now make a guess of a function of the type U/V = exp(F(x,y))

to get:
x d/dx F = y d/dy F +2

so you have x d/dx F -y d/dy F =2

d/dx (xF) - d/dy (yF) = 2

Don't see how to solve this, though.

p.s the derivatives above are partial btw.

 

I would rewrite this as

[tex]x \partial_x (\log U) - y \partial_y (\log U) - 1 = x \partial_x (\log V) - y \partial_y (\log V) + 1[/tex]

or

[tex]x \partial_x (\log U - \log V) - y \partial_y (\log U - \log V) = 2[/tex]

Define some function [itex]e^f = U/V[/itex] and you have

[tex]x \partial_x f - y \partial_y f = 2[/tex]

which should be easy to solve by characteristics. Mathematica gives

[tex]f = 2 \log x + C x y[/tex]

for arbitrary C. Then U and V can be any functions satisfying

[tex]\frac{U}{V} = x^2 e^{C x y}[/tex]

 

I don't know, I taught myself the method of characteristics when I started running across problems in my research that needed it.

 

Sometime I wonder why I even bother answering. :-/