Question to first order PDE with Characteristics Method We have the given problem
\begin{align} 
-2yu_{x} + u_{y} + 2yu &= 2y \tag 1 \\
u(1,y) &= 1 + e^{-1-2y^2} \tag 2
\end{align}
where $x > 0, y \in \mathbb{R^*}$. We can modify the formula $(1)$ to get
\begin{align}
-2yu_{x} + u_{y} &= 2y - 2yu \\
\implies -2yu_{x} + u_{y} &= 2y(1-u) \\
\implies -u_{x} + \frac{u_{y}}{2y} &= 1-u \\
\implies u_{x} - \frac{u_{y}}{2y} &= u-1
\end{align}
so our problem would be
\begin{align}
u_{x} - \frac{u_{y}}{2y} &= u-1 \tag 3 \\
u(1,y) &= 1 + e^{-1-2y^2} \tag 4
\end{align}
The method of characteristics gives
$$\dfrac{dx}{1} = -2y dy = \dfrac{du}{u-1}$$

Edited after @MatthewCassell suggestion
Now, we have
\begin{align}
\dfrac{dx}{1} =  -2y dy \implies 
 C_{1} &= y^2+x
\end{align}
and
\begin{align}
\dfrac{dx}{1} = \dfrac{du}{u-1} \implies u &= C_{2} e^x + 1 \\
\implies C_{2} &= \frac{u-1}{e^x}
\end{align}
The general solution of the PDE expressed in the form of implicit equation
$$C_2=G(C_1)$$
So $$u=e^x G(x+y^2)+1$$
Now, using the initial condition $u(1,y)=1+e^{-1-2y^2}$ gives
$$1+e^{-1-2y^2} = e G(1+y^2)+1 \implies G(1+y^2) = \frac{e^{-1-2y^2}}{e}$$
I get something strange (from what I've seen at least). Is this possible?

Solution
\begin{align}
-2yu_{x} + u_{y} &= 2y - 2yu \\
\implies -2yu_{x} + u_{y} &= 2y(1-u) \\
\implies -u_{x} + \frac{u_{y}}{2y} &= 1-u \\
\implies u_{x} - \frac{u_{y}}{2y} &= u-1
\end{align}
so our problem would be
\begin{align}
u_{x} - \frac{u_{y}}{2y} &= u-1 \tag 3 \\
u(1,y) &= 1 + e^{-1-2y^2} \tag 4
\end{align}
The method of characteristics gives
$$\dfrac{dx}{1} = -2y dy = \dfrac{du}{u-1}$$
Now, we have
\begin{align}
\dfrac{dx}{1} =  -2y dy \implies 
 C_{1} &= y^2+x
\end{align}
and
\begin{align}
\dfrac{dx}{1} = \dfrac{du}{u-1} \implies u &= C_{2} e^x + 1 \\
\end{align}
The general solution of the PDE expressed in the form of implicit equation
$$C_2=G(C_1)$$
So $$u=e^x G(x+y^2)+1$$
Now, using the initial condition $u(1,y)=1+e^{-1-2y^2}$ gives
$$1+e^{-1-2y^2} = e G(1+y^2)+1 \implies G(1+y^2) = \frac{e^{-1-2y^2}}{e}$$
Then,
\begin{align}
G(1+y^2) &= e^{-2-2y^2}\\
&=e^{-2(1+y^2)}
\end{align}
So we can see that
\begin{align}
G(x+y^2) &= e^{-2(x+y^2)}\\
\implies u&=e^x \cdot e^{-2(x+y^2)} +1\\
&=e^{-x-2y^2} +1
\end{align}
which satisfies the PDE and boundary condition
