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

second order and first degree

We have,

$A x^2+B y^2=1$            .....(i)

Differentiating w.r. to $x$, we get

$2 A x+2 B y \frac{d y}{d x}=0 \Rightarrow A x+B y \frac{d y}{d x}=0$    .......(ii)

Differentiating w.r. to $x$, we get

$A+B\left(\frac{d y}{d x}\right)^2+B y \frac{d^2 y}{d x^2}=0$       .....(iii)

Multiplying (iii) by $x$ and subtracting (ii) from it, we get

$x\left(\frac{d y}{d x}\right)^2+x y \frac{d^2 y}{d x^2}-y \frac{d y}{d x}=0$

Clearly, it is a second order and first degree differential equation.