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(D) Given the general solution: $A x^2+B y^2=1$ Differentiating with respect to $x$: $2 A x + 2 B y \frac{d y}{d x} = 0 \implies A x + B y \frac{d y}{

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

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(D) Given the general solution: $A x^2+B y^2=1$ Differentiating with respect to $x$: $2 A x + 2 B y \frac{d y}{d x} = 0 \implies A x + B y \frac{d y}{d x} = 0 \dots (i)$ Differentiating again with respect to $x$: $A + B \left[ \left( \frac{d y}{d x} ight)^2 + y \frac{d^2 y}{d x^2} ight] = 0 \dots (ii)$ From $(i)$,$A = -\frac{B y}{x} \frac{d y}{d x}$. Substituting this into $(ii)$: $-\frac{B y}{x} \frac{d y}{d x} + B \left( \frac{d y}{d x} ight)^2 + B y \frac{d^2 y}{d x^2} = 0$ Dividing by $B$ (assuming $B 
eq 0$): $-\frac{y}{x} \frac{d y}{d x} + \left( \frac{d y}{d x} ight)^2 + y \frac{d^2 y}{d x^2} = 0$ Multiplying by $x$: $-y \frac{d y}{d x} + x \left( \frac{d y}{d x} ight)^2 + x y \frac{d^2 y}{d x^2} = 0$ Thus,$x y \frac{d^2 y}{d x^2} + x \left( \frac{d y}{d x} ight)^2 - y \frac{d y}{d x} = 0$.

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

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