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The given differential equation is $\frac{dy}{dx} = \frac{x+y}{x}$.
This can be rewritten as $\frac{dy}{dx} = 1 + \frac{y}{x}$.
Let $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting into the differential equation, we get $v + x\frac{dv}{dx} = 1 + v$.
This simplifies to $x\frac{dv}{dx} = 1$, so $\frac{dv}{dx} = \frac{1}{x}$.
Integrating both sides with respect to $x$, we get $\int dv = \int \frac{1}{x} dx$, which gives $v = \ln|x| + C$.
Substituting back $v = \frac{y}{x}$, we have $\frac{y}{x} = \ln|x| + C$.
Thus, $y = x\ln|x| + Cx$.
Now, we use the initial condition $y(1) = 0$.
Substituting $x = 1$ and $y = 0$, we get $0 = 1\ln|1| + C(1)$, which simplifies to $0 = 0 + C$, so $C = 0$.
Therefore, the particular solution is $y = x\ln|x|$.
Correct Answer: y = x ln|x|