Question

Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis

Answer 1

According to question the given curve is
Since parabola has axis along positive 𝑦-axis, therefore, the equation is
$x^2=4ay$ (i)
$x^2=4ay$

Differentiating both sides w.r.t. x
we get,

$2x=4a\frac{dy}{dx}$

$x=2a\frac{dy}{dx}$

$x=2a{y}^{\prime }$ (ii)

Dividing $\left(ii\right)$ by $\left(i\right)$ then we get,

$\frac{x}{x^2}=\frac{2a{y}^{\prime }}{4ay}$

$\frac{1}{x}=\frac{y^{\prime }}{2y}$

$2y=x{y}^{\prime }$

$x{y}^{\prime }-2y=0$

Final Answer:
Hence, the required differential equation is

$x{y}^{\prime }-2y=0$