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

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Vishal Baghel | Contributor-Level 10

10 months ago

The equation of the parabola having the vertex at origin and the axis along the positive y-axis is:

$x^{2} = 4 a y$

Differentiating equation (1) with respect to x, we get:

$2 x = 4 a y'$

Dividing equation (2) by equation (1), we get:

$\begin{array}{l} \frac{2 x}{x^{2}} = \frac{4 a y^{'}}{4 a y} \\ \Rightarrow \frac{2}{x} = \frac{y^{'}}{y} \\ \Rightarrow x y^{'} = 2 y \\ \Rightarrow x y^{'} - 2 y = 0 \end{array}$

This is the required differential equation.

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dy/dx = 2y/ (xlnx).
dy/y = 2dx/ (xlnx).
ln|y| = 2ln|lnx| + C.
ln|y| = ln (lnx)²) + C.
y = A (lnx)².
(ln2)² = A (ln2)². ⇒ A=1.
y = f (x) = (lnx)².
f (e) = (lne)² = 1² = 1.