Find the general solution of xy' + y = ${\sqrt{x^2+y^2}}$
So, I've tried using u=$\frac{y}{x}$, and that gets me to a point where I have.
$\int \frac{1}{\sqrt{1+u^2}-2u}du$ = $\int\frac{1}{x}dx$
The right side I know how to solve, I just have no idea what to do with this integral on the left.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align} \int{\dd u \over \root{1 + u^{2}} - 2u} & = \int{\dd t \over t} - {2 \over 3}\int{2t \over t^{2} - 1/3}\,\dd t = \ln\pars{\verts{t}} - {2 \over 3}\ln\pars{\verts{t^{2} - {1 \over 3}}} \\[5mm] & = \ln\pars{\verts{\root{1 + u^{2}} - u}} - {2 \over 3}\ln\pars{\verts{\pars{\root{1 + u^{2}} - u}^{2} - {1 \over 3}}} \end{align} $$ = \bbx{\ln\pars{\verts{\root{1 + u^{2}} - u}} - {2 \over 3}\ln\pars{\verts{2u^{2} + {2 \over 3} -2u\root{1 + u^{2}}}} + \pars{~\mbox{a constant}~}} $$