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(B) Given the equation of the family of curves: $a x^2+b y^2=1$. Differentiating with respect to $x$, we get: $2 a x+2 b y \frac{d y}{d x}=0$, which s

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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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(B) Given the equation of the family of curves: $a x^2+b y^2=1$. Differentiating with respect to $x$, we get: $2 a x+2 b y \frac{d y}{d x}=0$, which simplifies to $a x+b y \frac{d y}{d x}=0$. From this, we have $a = -\frac{b y}{x} \frac{d y}{d x}$. Substituting $a$ into the original equation: $(-\frac{b y}{x} \frac{d y}{d x}) x^2 + b y^2 = 1 \Rightarrow -b x y \frac{d y}{d x} + b y^2 = 1$. Alternatively, differentiating $a x+b y \frac{d y}{d x}=0$ again with respect to $x$: $a + b \left( y \frac{d^2 y}{d x^2} + (\frac{d y}{d x})^2 ight) = 0$. Since $a = -\frac{b y}{x} \frac{d y}{d x}$, we substitute this: $-\frac{b y}{x} \frac{d y}{d x} + b y \frac{d^2 y}{d x^2} + b (\frac{d y}{d x})^2 = 0$. Dividing by $b$ (assuming $b 
eq 0$): $-\frac{y}{x} \frac{d y}{d x} + y \frac{d^2 y}{d x^2} + (\frac{d y}{d x})^2 = 0$. Multiplying by $x$: $-y \frac{d y}{d x} + x y \frac{d^2 y}{d x^2} + x (\frac{d y}{d x})^2 = 0$. Thus, the differential equation is $x y \frac{d^2 y}{d x^2} + x (\frac{d y}{d x})^2 - y \frac{d y}{d x} = 0$.

The differential equation corresponding to the family of curves given by $a x^2+b y^2=1$, where $a$ and $b$ are arbitrary constants, is:

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