Find $f\in C^{1}$ such that $x\partial_{x}f+y\partial_{y}f=(x^2+y^2)^{1/2}$ In a problem I am looking to find a function $C^{1}$, $f:\mathbb{R}^{2}\to\mathbb{R}$ such that
\begin{equation*}
    x\frac{\partial f}{\partial x}(x,y)+y\frac{\partial f}{\partial y}(x,y)=(x^2+y^2)^{1/2}
\end{equation*}
I tried doing the change of variables
\begin{eqnarray*}
    x=u\text{ and }y=uv
\end{eqnarray*}
getting
\begin{eqnarray*}
    u\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+uv\frac{\partial f}{\partial u}\frac{\partial u}{\partial x} & = & \sqrt{u^{2}+u^{2}v^{2}}\\
    \Rightarrow u\frac{\partial f}{\partial u} + u\frac{\partial f}{\partial u} & = & \sqrt{u^{2}+u^{2}v^{2}}\\
    \Rightarrow 2u\frac{\partial f}{\partial u} & = & |u|\sqrt{1+v^{2}}\\
\end{eqnarray*}
But from this point I don't know how to move forward, I tried to integrate u which led me to a resolution by trigonometric substitution (giving something of the form $f(x,y)=\frac{1}{2}\sec(\theta(x,y))+C$)
I would appreciate any ideas or indications on how to proceed.

$$x\frac{\partial f}{\partial x}(x,y)+y\frac{\partial f}{\partial y}(x,y)=(x^2+y^2)^{1/2}$$
Charpit-Lagrange characteristic ODEs :
$$\frac{dx}{x}=\frac{dy}{y}=\frac{df}{(x^2+y^2)^{1/2}}$$
A first characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{y}$ :
$$\frac{y}{x}=c_1$$
A second characteristic equation comes from solving $\frac{dx}{x}=\frac{df}{(x^2+y^2)^{1/2}}\quad\implies\quad \frac{dx}{x}=\frac{df}{(x^2+(c_1 x)^2)^{1/2}}$
$$f- x(1+(c_1)^2)^{1/2}=c_2$$
The general solution of the PDE expressed on implicit form $c_2=F(c_1)$ is :
$$f-x(1+(\frac{y}{x})^2)^{1/2}=F\left(\frac{y}{x}\right)$$
$$\boxed{f(x,y)=(x^2+y^2)^{1/2}+F\left(\frac{y}{x}\right)}$$
$F$ is an arbitrary function (to be determined according to some boundary conditions).
