Solve the PDE by the method of characteristics. I am trying to figure out where my solution went wrong. I am off by a factor of two.
$$ u_x + u_y + u = e^{x+2y}$$
I first found that the characteristic curves are determined by $$\frac{dy}{dx} = 1 \implies y-x = C.$$
I then solved the ODE $$\frac{du}{dx} + u = e^{x+2y}$$
I found $u = \frac12 e^{x+2y} + e^{-x}K(C)$ giving $$u = \frac{e^{x+2y}}{2} + e^{-x}F(y-x)$$ as the general solution where $F$ is an arbitrary function. Where does my work go wrong?
The end solution should be half of what it currently is.

$$\begin{align}
u_x + u_y + u &= e^{x + 2y} \\
\implies u_x + u_y &= e^{x + 2y} - u \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)\\
\end{align} $$
Setting $u = u(x(s),y(s))$ we find
$$\begin{align}
\frac{d}{ds} u &= \frac{\partial u}{\partial x} \cdot \frac{dx}{ds} + \frac{\partial u}{\partial y} \cdot \frac{dy}{ds} \\
&= \frac{\partial u}{\partial x} \cdot 1 + \frac{\partial u}{\partial y} \cdot 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \\
&= e^{x + 2y} - u
\end{align}$$
Where $(1)$ comes from our original PDE at $(*)$. Equating, we find
$$\begin{align}
\frac{dy}{ds} &= 1  \\
\frac{dx}{ds} &= 1 \implies \frac{dx}{dy} = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \\
\frac{du}{ds} &= e^{x + 2y} - u \implies \frac{du}{dy} + u = e^{x + 2y} \ \ \ \ \ \ \ \ \ \ \ \ \ (3) \\
\end{align}$$
Solving $(2)$ and $(3)$
$$x(y) = x_0 + y \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$$
$$\begin{align}
\frac{du}{dy} + u &= e^{x + 2y} \\
&= e^{x_0 + 3y} \\
\implies (e^{y}u)' &= e^{x_0 + 4y} \\
\implies e^{y}u &= \frac{e^{x_0 + 4y}}{4} + f(x_0) \\
\implies u &= e^{-y} \bigg(\frac{e^{x_0 + 4y}}{4} + f(x_0) \bigg) \\
&= \frac{e^{x_0 + 3y}}{4} + e^{-y}f(x_0) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5) \\
\end{align}$$
and using $(4) \implies x_0 = x - y$ we find
$$u = \frac{e^{x + 2y}}{4} + e^{-y}f(x - y)$$
