The differential equation whose solution is $A{x}^2+B{y}^2=1$, where $A$ and $B$ are arbitrary constants, is of:

Given: $A{x}^2+B{y}^2=1$. There are two arbitrary constants, so the order is $2$.Differentiate once: $2Ax+2By\frac{dy}{dx}=0⟹Ax+By{y}^{\prime }=0\dots (1)$ Differentiate again: $A+B(y^{\prime }{)}^2+By{y}^{\prime \prime }=0⟹A+B[(y^{\prime }{)}^2+y{y}^{\prime \prime }]=0\dots (2)$ From (1), $A=-\frac{By{y}^{\prime }}{x}$.Substitute into (2):$-\frac{By{y}^{\prime }}{x}+B[(y^{\prime }{)}^2+y{y}^{\prime \prime }]=0$ Dividing by $B$:$-\frac{y{y}^{\prime }}{x}+(y^{\prime }{)}^2+y{y}^{\prime \prime }=0⟹xy{y}^{\prime \prime }+x(y^{\prime }{)}^2-y{y}^{\prime }=0$ The highest order derivative is $y^{\prime \prime }$ (order 2) and its power is $1$ (degree 1).