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## Homework Statement

[tex] (x+y)u_{x} + yu_{y} = 0 [/itex]

[tex] u(1,y) = \frac{1}{y} + ln(y) [/itex]

## Homework Equations

...

## The Attempt at a Solution

Here was my approach via method of characteristics:

[tex] \frac{dy}{dt} = y \to y(t) = Se^{t} [/tex]

[tex] y(t=0) = S [/tex]

[tex] \frac{dx}{dt} = x+y [/tex]

[tex] x'-x = Se^{t} [/tex]

Via integrating factor we arrive at:

[tex] x(t) = Ste^{t} + g(s)e^{t} [/tex]

[tex] x(t=0) = 1 \to 1 = g(s) [/tex]

Thus we are left with:

[tex] x(t) = Ste^{t} + e^{t} [/tex]

and

[tex] y(t) = Se^{t} [/tex]

You can check for yourself that the following is true:

[tex] \frac{dU(x(t),y(t))}{dt} = (x+y)u_{x} + yu_{y} = 0 [/tex]

Thus our general solution is:

[tex] U(t) = f(s) [/tex]

A general function of s.

However, we have found that it is impossible to explicitly solve for s. Am I missing something?