Differential Equations — Class 12 Maths Solution

. : Equation of parabola having vertex at origin and axis along positive $y$-axis is

${x^2} = 4ay$, where $a$ is the parameter. …(1)

Differentiating (1) w.r.t. $x$,

we get
$2x = 4a{y_1} \Rightarrow \cfrac{{2x}}{{{y_1}}} = 4a$

…(2)
Substituting the value of $4a$ from equation (2) in equation (1),

we get
${x^2} = \cfrac{{2x}}{{{y_1}}}y \Rightarrow {x^2}{y_1} - 2xy = 0 \Rightarrow x{y_1} - 2y = 0$

which is the required differential equation.