# How can I solve the following PDE?

How to solve the following PDE

$x^2 v_x-y^2 v_y=0$ with $v\rightarrow e^x$ as $y\rightarrow\infty$

I found characteristic curves as $c_1=1/x+1/y$ and $v(x,y)=c_2$

and then ? Or is there anybody who solved it by another method?

Then you solve it via characteristic curves as usual. There's nothing special, except that your border condition is at infinity, but then again, what's so special about it? Your $v$ is essentially a function of ${1\over x}+{1\over y}$; as $y\to\infty$, the argument approaches $1\over x$, and the function approaches $e^x$. What function could it be, really?
• $$f\left({1\over x}\right)=e^x;\quad f(x)=?$$ – Ivan Neretin Apr 12 '17 at 20:48
• It is $e^{x^{(-1)}}$. – HD239 Apr 12 '17 at 21:09
• That's right; now apply that function to ${1\over x}+{1\over y}$. The result is your $v(x,y)$. – Ivan Neretin Apr 12 '17 at 21:11
• General solution of equation is $$v=f\left(\frac1x+\frac1y\right)$$
• $f\left(\frac1x\right)=e^x\;\Rightarrow\; f(x)=e^{\frac1x}$
• Answer: $$v=e^\left(\frac{xy}{x+y}\right)$$