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I was trying to solve the following problem:

The integral surface of the pde $xu_x+yu_y=0$ satisfying the condition $u(1,y)=y$ is given by:

(a)$u(x,y)=y/x,$
(b)$u(x,y)=2y/(x+1),$
(c)$u(x,y)=y/(2-x),$
(d)$u(x,y)=y+x-1.$

I was trying to apply Lagrange's method but i could not get the desired result. I see that option $(a)$ satisfies the given equation and so $(a)$ should be the right choice. But i could not get it directly. Please help.Thanks in advance for your time.

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For PDEs of the form $a(x,y)u_x+b(x,y)u_y=0$, use the method of characteristic curves.

Think of the PDE as $\nabla u\cdot (a(x,y),b(x,y))=0$ so that, geometrically it must be that ${dy\over dx}={b(x,y)\over a(x,y)}$. The solution of the ODE defines the characteristic curves of the PDE, along which solutions (to the PDE) are constant. Then $u(x,y)=f(C)$ where the function $f$ is determined from the given auxiliary condition.

For example, here the ODE is ${dy\over dx}={y\over x}$ so $C=\ln|y/x|$. Then $u(x,y)=f(C)=f(\ln|y/x|)$. To determine $f$, apply the given condition: $u(1,y)=f(\ln|y|)=y\implies f(y)=e^y$. Thus, $u(x,y)=\exp(\ln|y/x|)=|y/x|$, corresponding to your answer (a).

(Maybe you were given $x,y>0$?)

Here are some more examples of this technique.

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  • $\begingroup$ thanks a lot sir for elaborate explanation. $\endgroup$ – user52976 Dec 15 '12 at 16:05
  • $\begingroup$ Glad to help. Not saying this for my own benefit, but to benefit the site and participants in general. If you see an answer (or question) that was helpful, interesting, insightful, etc. please up vote it by clicking on the arrow above the number to the top left of it. $\endgroup$ – JohnD Dec 15 '12 at 16:07
  • $\begingroup$ Thanks a lot sir. I would keep your advice in mind. +1 from me. $\endgroup$ – user52976 Dec 15 '12 at 16:43
  • $\begingroup$ Actually, $u$ need not be of the form $f(\ln|y/x|)$. This use of absolute value forces $u(1,1)=u(1,-1)$, which need not be the case since any degree-$0$ homogeneous function solves the PDE. $\endgroup$ – user147263 Jun 3 '15 at 20:09
  • $\begingroup$ (Commenting now, because this came up in a similar question). $\endgroup$ – user147263 Jun 3 '15 at 20:16

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