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

second order and first degree

The correct answer is option (4) : second order and first degree

We have,

$Ax^2 + By^2=1$...............(i)

Differentiating w.r. to x, we get

$2Ax + 2By \frac{dy}{dx} = 0 ⇒Ax+ By \frac{dy}{dx} = 0 $ ..............(ii)

Differentiating w.r. to x, we get

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

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

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

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