Confusion solving linear first order PDE In my course, it has been presented to us that in order to solve a PDE of the form:
$$a(x,y)U_x +b(x,y)U_y=c(x,y)U+d(x,y)$$
The general method is to solve $$\frac{dy}{dx}=\frac{b(x,y)}{a(x,y)}$$ the equation of the characteristic curve. Then we have to solve:
$$\frac{d}{dx}U(x,y)=\frac{c(x,y)}{a(x,y)}U+\frac{d(x,y)}{a(x,y)}$$
I am struggling with a particular example:
$$U_x+yU_y=0 $$
In my lecture notes the first step is to solve $$\frac{dy}{dx}=\frac{y}{1}$$ and this obviously yields $y=Ce^x$ for some $C \in \mathbb{R}$, this is fine.
Solving the second equation we find that $$\frac{d}{dx}U(x,y)=0$$ Now here is where I am confused, in my notes the following is written after this point: "Now, observe that for $\frac{d}{dx}U(x,Ce^x)=U_x+Ce^xU_y=U_x+yU_y=0$ Thus, along each characteristic curve the solution is a constant and the solution can only
depend on C —that is $U(x,y)=f(C)$."
I really do not understand what this is saying and how this conclusion is reached, can anyone explain this further?
 A: Assume $a\not\equiv 0$, so that the PDE reads
$$
U_x + \tfrac{b}{a} U_y = \tfrac{c}{a} U + \tfrac{d}{a} ,
$$
with $a$, $b$, $c$, $d$ function of $(x,y)$.
Now, consider curves described by an equation of the form $y = y(x)$ in $y$-$x$ coordinates. Thus, the 'total' directional derivative of $U = U(x, y(x))$ along these curves satisfies $$
\tfrac{\text d}{\text d x} U = U_x + \tfrac{\text d y}{\text d x} U_y .
$$
according to the multivariate chain rule. The method of characteristics consists in setting consistently
$$
\tfrac{\text d}{\text d x} y = \tfrac{b}{a}, \qquad \tfrac{\text d}{\text d x} U = \tfrac{c}{a} U + \tfrac{d}{a}
$$
to transform the initial PDE into ordinary differential equations of the variable $x$.
Example. Set $a\equiv 1$, $b = y$, $c\equiv 0$, $d\equiv 0$. As explained above, the method of characteristics amounts to setting $y = y(x)$ with
$$
\tfrac{\text d}{\text d x} y = y, \qquad \tfrac{\text d}{\text d x} U = 0 .
$$
Thus, we know that $y = c_1 \text{e}^x$ and $U = c_2$ with $c_1$, $c_2$ arbitrary constants. Let us go back to our assumptions on $U$ to note that $U(x, c_1 \text{e}^x) = c_2$, which implicitly links the constants $c_1$, $c_2$. More precisely, let us assume that for a given fixed $x_0$, we know the expression of $U(x_0, c_1\text{e}^{x_0}) = f(c_1)$ for some suitable function $f$. It then follows that $c_2 = f(c_1)$, and thus $$U(x,y) = f(y\text{e}^{-x}).$$ See also this post for further attempts at explaining this fact.
