Q:

Solve First Order Linear PDE with Method of Characteristic

Given a problem
$x^2\,u_x+xy\,u_y=y\,u$
Then, the characteristic equation is
$\dfrac{dx}{x^2}=\dfrac{dy}{xy}=\dfrac{du}{yu}$
From the first two of those equation, i got
$C_1=\dfrac{y}{x}$
How to obtain the other $C_2$? So that i have implicit solution $F(C_1)=C_2$?

A:

$$\dfrac{dx}{x^2}=\dfrac{dy}{xy}=\dfrac{du}{yu}$$
From the first two ratio 
\begin{align}
\dfrac{dx}{x^2} &= \dfrac{dy}{xy} \\
\implies \dfrac{dx}{x} &= \dfrac{dy}{y} \\
\implies \log x - \log y &= \log c_1 \\
\implies\frac{x}{y} &= c_1
\end{align}
From the last two 
\begin{align}
\dfrac{dy}{xy} &= \dfrac{du}{yu} \\
\implies \dfrac{dy}{x} &= \dfrac{du}{u} \\
\implies \dfrac{dy}{c_1y} &= \dfrac{du}{u} \\
\implies \log y &= c_1(\log u - \log c_2) \\
\implies \log y &= \frac{x}{y}\log \left(\frac{u}{c_2} \right) \\
\implies \frac{u}{c_2} &= y^{\frac{y}{x}} \\
\implies c_2 &= uy^{-\frac{y}{x}}
\end{align}
So the  implicit solution is $$F \left(\frac{x}{y} \right)=uy^{-\frac{y}{x}}$$

