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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.

1 Answers 1

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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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    (Commenting now, because this came up [in a similar question](http://math.stackexchange.com/questions/1309608/method-of-characteristics-for-the-pde-xu-x-y-u-y-0-with-an-initial-conditio?lq=1)).2015-06-03