Any straight lines at constant distance of $P$ units from the origin is given by $x\cos \alpha +y\sin \alpha =P$

Differentiating both the sides w.r.t. $x$, we get

$\cos \alpha +\frac{dy}{dx}\sin \alpha =0$

i.e. $\frac{dy}{dx}=-\cot \alpha \Rightarrow \cos \alpha =-\frac{\frac{dy}{dx}}{\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}, \sin \alpha =\frac{1}{\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}$

$x\frac{\left(-{\displaystyle \frac{dy}{dx}}\right)}{\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}+y\frac{1}{\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}=P$

$\Rightarrow -x\frac{dy}{dx}+y=P\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$

Squaring both sides, we get

$x^2{\left(\frac{dy}{dx}\right)}^2-2xy\frac{dy}{dx}+y^2=P^2\left(1+{\left(\frac{dy}{dx}\right)}^2\right)$

Hence, order of the differential equation is $1$ and degree is $2$.

Therefore, $l{m}^2+m{l}^2=4+2=6$