The differential equation of the family of cu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The length of the normal to a curve is given by the formula $L = |y|\sqrt{1 + (\frac{dy}{dx})^2}$.Given that the length of the normal is a constan

Vedclass · Accepted Answer

${\left( {y\frac{{dy}}{{dx}}} \right)^2} = {k^2} - {y^2}$

Vedclass · Answer

(B) The length of the normal to a curve is given by the formula $L = |y|\sqrt{1 + (\frac{dy}{dx})^2}$.Given that the length of the normal is a constant $k$,we have $|y|\sqrt{1 + (\frac{dy}{dx})^2} = k$.Squaring both sides,we get $y^2(1 + (\frac{dy}{dx})^2) = k^2$.Expanding the equation,we get $y^2 + y^2(\frac{dy}{dx})^2 = k^2$.Rearranging the terms,we obtain $y^2(\frac{dy}{dx})^2 = k^2 - y^2$,which can be written as $(y\frac{dy}{dx})^2 = k^2 - y^2$.

The differential equation of the family of curves for which the length of the normal is equal to a constant $k$,is given by

Similar Questions

The curve $y=(\cos x+y)^{1 / 2}$ satisfies the differential equation

If $a$ and $b$ are arbitrary constants,then the differential equation corresponding to the family of curves $y = \tan(ax + b)$ is:

Let $c_1, c_2, c_3, c_4$ be arbitrary constants. The order of the differential equation,corresponding to $y=c_1 e^x+c_2 e^{\log _{e} x}+c_3 \sin ^2 x-c_4\left(\cos ^2 x-1\right)$ is

The normal form of the equation of a line is $x \cos \alpha + y \sin \alpha = p$. Find the differential equation of the family of all such lines, where $p$ and $\alpha$ are arbitrary constants.

Form the differential equation of the family of lines $y = mx + \frac{4}{m}$ by eliminating the arbitrary constant $m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module