# I have a first order PDE: x u x </msub> + ( x + y )

I have a first order PDE:
$x{u}_{x}+\left(x+y\right){u}_{y}=1$
With the initial condition:
I have calculated result in Mathematica: $u\left(x,y\right)=\frac{y}{x}$, but I am trying to solve the equation myself, but I had no luck so far. I tried with method of characteristics, but I could not get the correct results. I would appreciate any help or maybe even whole procedure.
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Hailee Henderson
$x{u}_{x}+\left(x+y\right){u}_{y}=1$
${u}_{x}+\left(\frac{y}{x}+1\right){u}_{y}=\frac{1}{x}$
$\frac{dx}{dt}=1$, letting x(1)=1 , we have x=t
$\frac{dy}{dt}=\frac{y}{x}+1=\frac{y}{t}+1$, letting $y\left(1\right)={y}_{0}$, we have
$y\left(1\right)={y}_{0}$
$\frac{du}{dt}=\frac{1}{x}=\frac{1}{t}$, letting $u\left(1\right)=f\left({y}_{0}\right)$, we have
$u\left(x,y\right)=\mathrm{ln}t+f\left({y}_{0}\right)=\mathrm{ln}x+f\left(\frac{y}{x}-\mathrm{ln}x\right)$
$u\left(1,y\right)=y$:
$f\left(y\right)=y$
$\therefore u\left(x,y\right)=\mathrm{ln}x+\frac{y}{x}-\mathrm{ln}x=\frac{y}{x}$