 Solution

The two affine lines meet at $(a,b)=(2,1)$. Translating their intersection to the origin, $X=x-2$, $Y=y-1$, removes both constant terms and gives the homogeneous first-order differential equationOn a patch with $X\ne 0$, put $Y=uX$. Then $Xdu/dX=(u^2+4)/(2-u)$, and separation of variables givesThus one local real implicit form isThe arctangent branch must be continued on the chosen solution patch. More geometrically, set $r=\sqrt{4X^2+Y^2}$ and take a continuous local argument $\theta =\arg (2X+iY)$; then $\theta -\ln r=C$. This form extends through $X=0$ where the original slope is defined. The intersection itself is excluded because the original right-hand side is $0/0$ there, and points with $2X-Y=0$ do not define a finite derivative. No real constant-$u$ solution was lost, because $u^2+4$ has no real zero. Translating the intersection works precisely because the two affine linear forms then become homogeneous linear forms.