# How to solve: $xu_{x}+u_{y}=1,u(x,0)=2\ln(x)$, $x>1.$

I have been trying to solve the following problem:

Let $u(x,y)$ be the solution to the Cauchy Problem $$xu_{x}+u_{y}=1,\;\;u(x,0)=2\ln(x),\quad x>1.$$ Then $u(e,1)=?$ I was trying to solve it by Lagrange's method but could not progress. Can someone point me in the right direction? (A certain property or theorem that I have to use to find out $u(x,y)$.)

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$u(x,y)=2ln(x)-y$. – ougao Dec 14 '12 at 18:56
How do you get that? Can you refer to the formula used to get it? – learner Dec 14 '12 at 19:17
Just some attempt, I can not gurantee the solution is complete, note that given such problem, the first simple idea is to guess $u=2ln(x)+f(y)$, it is lucky this type really gives us a solution here. – ougao Dec 14 '12 at 19:20
Thanks a lot @ougao. Your input has been useful. – learner Dec 14 '12 at 19:25

Using the method of characteristics, let $u(x,y) = u(x(s),y(s)) = f(s)$. Then $$\dfrac{df}{ds} = \dfrac{\partial u}{\partial x} \dfrac{dx}{ds} + \dfrac{\partial u}{\partial y} \dfrac{dy}{ds} = x'(s) u_x +y'(s) u_y = xu_x + u_y$$ Hence, let us set $x'(s) = x$ and $y'(s) = 1$. This gives us $$x(s) = c_1 e^s; \,\,\,\,\,\,\,\, y(s) = s \,\,\,\,\,\,\, f'(s) = 1 \implies f(s) = s + f(0)$$ We also have that $u(x,0) = 2 \ln(x)$. $y=0 \implies s = 0$. Hence, we have that $$f(0) = 2 \ln(x(0)) = 2 \ln(c_1) = f(0) \implies f(0) = 2 \ln(c_1)$$ Hence, $$f(0) = 2 \ln(xe^{-y}) \implies u(x,y) = s + 2 \ln(xe^{-y})$$ $$= y + 2 \ln(x) - 2y$$ $$= 2\ln(x) - y$$
You left out the $s$ term at the end so it's $2\ln(x) - y$, I'll edit it... – Zarrax Dec 14 '12 at 21:07