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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?

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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?

Alternatively, you may abandon this approach (despite its being 100% applicable and 75% completed) and switch to some other method, of which there are plenty.

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  • $\begingroup$ I don' understand completely. Can you help more pls. for this method (or another method)? $\endgroup$ – HD239 Apr 12 '17 at 20:47
  • $\begingroup$ $$f\left({1\over x}\right)=e^x;\quad f(x)=?$$ $\endgroup$ – Ivan Neretin Apr 12 '17 at 20:48
  • $\begingroup$ It is $e^{x^{(-1)}}$. $\endgroup$ – HD239 Apr 12 '17 at 21:09
  • $\begingroup$ That's right; now apply that function to ${1\over x}+{1\over y}$. The result is your $v(x,y)$. $\endgroup$ – Ivan Neretin Apr 12 '17 at 21:11
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  • General solution of equation is $$v=f\left(\frac1x+\frac1y\right)$$
  • $f\left(\frac1x\right)=e^x\;\Rightarrow\; f(x)=e^{\frac1x}$
  • Answer: $$v=e^\left(\frac{xy}{x+y}\right)$$
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