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(B) The general equation of a parabola with its axis along the $y$-axis is given by $x^2 = 4ay$, where $a$ is an arbitrary constant.Step $1$: Differen

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$x \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$

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(B) The general equation of a parabola with its axis along the $y$-axis is given by $x^2 = 4ay$, where $a$ is an arbitrary constant.Step $1$: Differentiate both sides with respect to $x$: $2x = 4a \frac{dy}{dx}$.Step $2$: From this, we get $a = \frac{2x}{4(dy/dx)} = \frac{x}{2(dy/dx)}$.Step $3$: Substitute the value of $a$ back into the original equation: $x^2 = 4 \left( \frac{x}{2(dy/dx)} \right) y$.Step $4$: Simplify the equation: $x^2 = \frac{2xy}{dy/dx}$.Step $5$: Rearrange to get $x(dy/dx) = 2y$, or $x \frac{dy}{dx} - 2y = 0$. Note: The standard form for parabolas with axis as $y$-axis is $y = ax^2 + b$. If the vertex is at the origin, it is $x^2 = 4ay$. Given the options, the intended family is $x^2 = 4ay$. Differentiating $x^2 = 4ay$ gives $2x = 4a y'$, so $a = x/(2y')$. Substituting into $x^2 = 4ay$ gives $x^2 = 4(x/2y')y$, which simplifies to $x y' = 2y$. Since this is not in the options, we re-evaluate the family $y = ax^2 + b$. Differentiating twice gives $y'' = 2a$. This does not match. Re-evaluating $x^2 = 4ay$: $2x = 4ay' \implies x = 2ay'$. Differentiating again: $1 = 2ay''$. Thus $2a = 1/y''$. Substituting $2a$ into $x = 2ay'$ gives $x = (1/y'')y'$, which leads to $x y'' - y' = 0$.

The differential equation of the family of all parabolas whose axis is the $y$-axis is ...

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