Lost solution to PDE? I’m solving the PDE for function $u(x,y)$ with parameter $\lambda$
\begin{align}
\lambda uu_x+u_y=0.
\end{align}
Using the method of characteristics,
\begin{align}
\frac{\mathrm du}{\mathrm ds}=0&\rightarrow u=\Phi(C),\\
\frac{\mathrm dx}{\mathrm ds}=\lambda u&\rightarrow x=\lambda su+C,\\
\frac{\mathrm dy}{\mathrm ds}=1&\rightarrow y=s.
\end{align}
Combining we find that
\begin{align}
u=\Phi\left(x-\lambda yu\right),
\end{align}
which agrees with what Wolfram Alpha tells me. The problem is this doesn’t work for all solutions, specifically the case $u=x/y$, which upon substitution into the PDE you’ll see works for $\lambda=1$, but fails to yield an identity when plugged into the general solution.
I’ve tried taking the different initial conditions $(x_0,y_0)=(0,C)$, but that solution  has the same issue. Could someone enlighten me as to how I’m losing this solution? Is there an assumption I am taking at some point to do so?
 A: The short answer: the way you have solved it assumes the initial conditions can be specified along $y=0$. But this is not possible for $u=x/y$ as the function is not defined here.
Longer answer: Lets review the whole idea behind the method of characteristics: we specify the initial conditions of the PDE along a curve (for example the $x$-axis) and then find characteristic lines where the function stays constant (or vary in a way we can compute). Then to find $u(x,y)$ we simply follow the characteristic lines from $(x,y)$ till it hits the curve at $(x_0,y_0)$ and then we can just read off the function value $u(x,y)$ from the initial conditions $u(x_0,y_0)$. Notice that there are in general two parameters, $x_0$ and $y_0$, that specify the point on the initial curve. Often we make the simplification that the initial conditions are set on a simple line, say the $x$-axis, and then it reduces to a single parameter (your $C$).
The way you have solved the equation, by taking $\frac{dy}{ds}=1\to y=s$ instead of $y=s+y_0$, you have implicitly assumed that the initial conditions of the PDE can be specified along $y=0$ (for $s=0$ with $x_0 = C$). This is not possible for $u=x/y$. For $u = x/y$ the function is divergent here and as a consequence the characteristic curves you find all pass through $x=y=0$ so if $u(0,0)$ was set to be finite this would mean $u$ has to be constant for this to be consistent (and this is a solution).
If we go back and do it more carefully, we find
$$u(s) = u_0 = f(x_0,y_0)$$
$$x(s) = su_0+x_0$$
$$y(s) = s+y_0$$
where $f(x_0,y_0)$ is our initial condition.
We have seen that taking $y_0=0$ is not allowed if we want a solution like $u=x/y$. If we instead imagine the initial conditions are specified along the curve $y=1$ (i.e. $y_0=1$) then $x(s) = (y-1)u + x_0$ so $u = F(x-(y-1)u)$ for some function $F(t) = f(t,1)$.
Now plugging in $u=x/y$ (i.e. $u(x_0,1) = x_0$ so the initial conditions is $F(x)=x$) we find $x/y = F(x/y)$ which is exactly what we expect.
