1. We are given the family of solutions

$Ax^2 + By^2 = 1$

where $A$ and $B$ are arbitrary constants.

Since there are two arbitrary constants, the required differential equation should be of order 2.

2. Differentiate once with respect to $x$:

$\frac{d}{dx}(Ax^2 + By^2)=0$

$2Ax + 2By\frac{dy}{dx}=0$

$Ax + Byy' = 0 \quad \text{...(1)}$

3. Differentiate again with respect to $x$:

Differentiate $Ax + Byy' = 0$:

$A + B\frac{d}{dx}(yy')=0$

Using product rule,

$\frac{d}{dx}(yy') = (y')^2 + yy''$

So,

$A + B\big((y')^2 + yy''\big)=0 \quad \text{...(2)}$

4. Eliminate $A$ and $B$.

From the original equation,

$Ax^2 + By^2 = 1 \quad \text{...(3)}$

From (1),

$A = -\frac{Byy'}{x}$

Substitute into (2):

$-\frac{Byy'}{x} + B\big((y')^2 + yy''\big)=0$

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

Ignoring the trivial constant factor $B$, we get

$yy'' + (y')^2 - \frac{yy'}{x} = 0$

This is the required differential equation.

5. Determine order and degree.

• Highest order derivative present is $y''$. So the order = 2.
• The highest order derivative $y''$ appears to the first power. So the degree = 1.

6. Hence, the differential equation is of

$\boxed{\text{second order and first degree}}$

So the correct option is D.