Question

Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of $x$-axis.

Answer 1

According to question the given curve is
Since parabola has axis along positive $x$-axis,
Therefore, the equation is $y^2=4ax$

$y^2=4ax$

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

$\frac{d\left(y^2\right)}{dx}=\frac{d\left(4ax\right)}{dx}$

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

$y^2=4ax$

Putting

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

$y^2=2y\frac{dy}{dx}\times x$

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

Final Answer:
Hence, the required differential equation is

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