Method of characteristics. Small question about initial conditions.

Here is a technique you can follow. We have the equations

$$ \frac{dx}{dt}=x, \frac{dy}{dt}=x+y, \frac{du}{dt}=1 \longrightarrow (*). $$

From the first two equations, we get

$$ \frac{dy}{dx}=\frac{x+y}{x} \implies y = x\ln(x) + c x \implies c=\frac{y-x\ln(x)}{x}\longrightarrow (1) .$$

Now, the first and the third equations in $(*)$ give

$$ \frac{du}{dx}=\frac{1}{x}\implies u(x,y)= \ln(x)+f(c) \longrightarrow (2).$$

Using $(1)$, $(2)$ becomes

$$\implies u(x,y)=\ln(x)+f\left(\frac{y-x\ln(x)}{x}\right) \longrightarrow (**). $$

Now, we exploit the initial condition in $(**)$ to find the function $f$

$$ u(1,y) = y = 0+f(y) \implies f(y)=y. $$

Substituting back in $(**)$, we have

$$ u(x,y)=\ln(x)+\left(\frac{y-x\ln(x)}{x}\right)$$

$$ u(x,y) = \frac{y}{x}. $$

The idea with method of characteristics is there is a initial condition variable $s$ that parametrizes the initial curve, and a characteristic variable $t$ that dictates the "flow" of the characteristics away from the initial curve.

In your problem, your initial condition is $u(x=1,y) = y$, so the initial curve is the curve given by $x = 1$, which can be parametrized by $\{(1,s): s \in \mathbb{R}\}$. So this gives us $x(0,s) = 1, y(0,s) = s, u(0,s) = s$.

Solving the ODEs, $$ x(t,s)=e^t,\;\; y(t,s)=e^t(t+s),\;\; z(t,s)=t+s $$ Then $$ u(x,y)=t+s=\frac{e^t(t+s)}{e^t}=\frac{y}{x} $$