JEE Main · 2006 · 30 Apr 2006 · Offline · Differential Equations

The solution $A{x}^2+B{y}^2=1$ carries two arbitrary constants, so removing them requires two differentiations and yields a second-order equation. Differentiating once gives $Ax+By{y}^{\prime }=0$, and differentiating again gives $A+B{y}^{\prime 2}+By{y}^{\prime \prime }=0$. Eliminating $A$ and $B$ between these relations leaves $xy{y}^{\prime \prime }+x{y}^{\prime 2}-y{y}^{\prime }=0$. The highest derivative present is $y^{\prime \prime }$, so the order is two, and it appears only to the first power, so the degree is one; the equation is of second order and first degree.