$1$st order homogenous equation $xdy = (y -\sqrt {x^2 + y^2}) dx$

Question

$x\,dy = (y -\sqrt{x^2 + y^2})\,dx$

If I substitute $z(x)x$ over $y(x)$, where $z(x)$ is a new function,

then $dy = z\,dx + x\,dz$

I am doubting the way I am doing this is correct...

if $y(x) = z(x)x$ then the original equation would be $zy(z\,dx + x\,dz) = (xz -\sqrt{x^2 + x^2 z^2})\,dx$

Please help me to solve it

Answer 1

Upon rearrangement, the terms cancel out, so that the integral is easier to solve. Take the x inside the square root, and express the terms in the form of y/x. Now, if you substitute y = (v)×(x), you'll see that it reduces to [dv/ √(1+v^2)] = -dx/x , and I hope you can take it from here.

Answer 2

$\newcommand{\d}{\mathrm{d}}$ Write $$\begin{align} \alpha &= x \d y-y\d x + r \d x\\ \beta &= x^2+y^2-r^2\\ \gamma &= \d r+\d y \end{align}$$ Then $$\begin{align} 2x\alpha-2\d y\beta-(r-y)\d \beta &= 2r(r-y)\gamma \\ 2(r+y)\alpha+2\d x\beta-x\d \beta &= 2rx\gamma \end{align}$$ so that, if $r\neq 0$ and either $x\neq 0$ or $r\neq y$, then $\alpha\equiv 0$ and $\beta\equiv 0$ imply $\gamma\equiv 0$.