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Question

(A) The equation of a parabola with vertex at the origin and axis along the positive $Y$-axis is given by $x^2 = 4ay$,where $a > 0$ is an arbitrary co

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

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(A) The equation of a parabola with vertex at the origin and axis along the positive $Y$-axis is given by $x^2 = 4ay$,where $a > 0$ is an arbitrary constant. To find the differential equation,we differentiate with respect to $x$: $2x = 4a \frac{dy}{dx}$ $\Rightarrow a = \frac{2x}{4(dy/dx)} = \frac{x}{2(dy/dx)}$. Substituting the value of $a$ back into the original equation: $x^2 = 4 \left( \frac{x}{2(dy/dx)} ight) y$ $x^2 = \frac{2xy}{dy/dx}$ $x^2 \frac{dy}{dx} = 2xy$ Dividing by $x$ (since $x 
eq 0$): $x \frac{dy}{dx} = 2y$ $x \frac{dy}{dx} - 2y = 0$.

The differential equation of all parabolas having vertex at the origin and axis along the positive $Y$-axis is

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