Find the general integral of the homogeneous equation…

Find the general integral of the homogeneous equation $xy'-y=\sqrt{x^2+y^2}$.

Step 1: Substitute $y=vx$

With $y=vx$, $y'=v+xv'$, and for $x>0$ the equation becomes $xv+x^2v'-vx=x\sqrt{1+v^2}$, i.e.

\[x\frac{dv}{dx}=\sqrt{1+v^2}.\]

Step 2: Separate and integrate

\[\frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x} \quad\Longrightarrow\quad \ln\!\left(v+\sqrt{1+v^2}\right)=\ln|x|+C \quad\Longrightarrow\quad v+\sqrt{1+v^2}=Kx.\]

Step 3: Solve for $v$ and back-substitute

Isolating and squaring: $\sqrt{1+v^2}=Kx-v \Rightarrow 1=K^2x^2-2Kxv \Rightarrow v=\dfrac{K^2x^2-1}{2Kx}$. With $y=vx$ and $C=1/K$, this simplifies to

\[\boxed{\,x^2-2Cy-C^2=0.}\]