Form the differential equation of the family of ellipses having foci on the $y$-axis and centre at the origin.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The equation of the family of ellipses having foci on the $y$-axis and the centre at the origin is given by:
$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$ --- $(1)$
Differentiating equation $(1)$ with respect to $x$,we get:
$\frac{2x}{b^{2}}+\frac{2yy'}{a^{2}}=0$
$\Rightarrow \frac{x}{b^{2}}+\frac{yy'}{a^{2}}=0$ --- $(2)$
Dividing by $x$,we get $\frac{1}{b^{2}} = -\frac{yy'}{a^{2}x}$.
Differentiating equation $(2)$ with respect to $x$ again:
$\frac{1}{b^{2}}+\frac{y' \cdot y' + y \cdot y''}{a^{2}} = 0$
Substituting $\frac{1}{b^{2}} = -\frac{yy'}{a^{2}x}$ into the equation:
$-\frac{yy'}{a^{2}x} + \frac{(y')^{2} + yy''}{a^{2}} = 0$
Multiplying by $a^{2}x$:
$-yy' + x(y')^{2} + xyy'' = 0$
Thus,the required differential equation is:
$xyy'' + x(y')^{2} - yy' = 0$

Explore More

Similar Questions

The differential equation of all circles passing through the origin and having their centers on the $x$-axis is

The differential equation for all the straight lines which are at a unit distance from the origin is

Difficult
View Solution

The differential equation satisfied by the family of curves $y = ax \cos \left( \frac{1}{x} + b \right)$,where $a$ and $b$ are parameters,is

The differential equation corresponding to all the circles lying in the first quadrant and touching the coordinate axes is

The differential equation of the family of circles whose center lies on the $x$-axis is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo