Find the solution to a PDE with an initial condition 
Find the solution to $u_x + y u_y = u$ with initial condition $u(0,y) = \cos(y)$.

Attempted solution - Suppose we parametrize a curve $(x,y)$ by a parameter $\xi$. So that
$$
u=u(x(\xi),y(\xi))
$$
$$
\frac{\mathrm{d}u}{\mathrm{d}\xi}=\frac{\mathrm{d}x}{\mathrm{d}\xi}\frac{\partial u}{\partial x}+\frac{\mathrm{d}y}{\mathrm{d}\xi}\frac{\partial u}{\partial y}
$$
Comparing to our original PDE
$$
u = \frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}
$$
We then set 
$$
\frac{\mathrm{d}u}{\mathrm{d}\xi}=u
$$
$$
\frac{\mathrm{d}x}{\mathrm{d}\xi}=1
$$
$$
\frac{\mathrm{d}y}{\mathrm{d}\xi}=y
$$
We then have 
$$
\mathrm{d}\xi=\frac{\mathrm{d}u}{u}= \mathrm{d}x = \frac{\mathrm{d}y}{y}
$$
Integrating we have $$\xi = \ln(u) = x = \ln(y) + C$$
Now for the general solution we have $$\frac{\mathrm{d}u}{\mathrm{d}\xi} = u \Rightarrow \mathrm{d}\xi = \frac{\mathrm{d}u}{u}\Rightarrow \xi = \ln(u) + C \Rightarrow u = Ce^{\xi}$$
The initial condition implies...
I am following the structure from another user that answered a similar question I had but I am now a bit lost applying the initial condition. I do know that $x = \xi$ but I am not sure how to get the rest of the solution from there.

$$u_x+yu_y=u$$
Set of characteristic ODEs: $\quad \frac{dx}{1}=\frac{dy}{y}=\frac{du}{u}$
First family of characteristics curves, from $\frac{dx}{1}=\frac{dy}{y} \quad\to\quad ye^{-x}=c_1$
Second family of characteristics curves, from $\frac{dy}{y}=\frac{du}{u} \quad\to\quad \frac{u}{y}=c_2$
General solution, with any differentiable fonction $F$ :
$$\frac{u}{y}=F(ye^{-x}) \quad\to\quad u=y\:F(ye^{-x}) $$
Condition : $\quad u(0,y)=\cos(y)=y\:F(ye^{0}) \quad$ determines the function $F$ :
$$F(t)=\frac{\cos(t)}{t} \quad \text{any } t\neq 0$$
Bringing this function $F$ into the above general solution gives 
$\quad u=y\:\frac{\cos(ye^{-x}) )}{ye^{-x}}$
$$u(x,y)=e^x\cos(y\:e^{-x})$$
