Differential Equations Question 220
Question: The differential equation of the family of curves for which the length of the normal is equal to a constant k, is given by
[Pb. CET 2004]
Options:
A) $ y^{2}\frac{dy}{dx}=k^{2}-y^{2} $
B) $ {{( y\frac{dy}{dx} )}^{2}}=k^{2}-y^{2} $
C) $ y{{( \frac{dy}{dx} )}^{2}}=k^{2}+y^{2} $
D) $ {{( y\frac{dy}{dx} )}^{2}}=k^{2}+y^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
The length of normal is given by, $ y\sqrt{1+{{( \frac{dy}{dx} )}^{2}}} $
$ \therefore $ $ y\sqrt{1+{{( \frac{dy}{dx} )}^{2}}}=k $
Therefore $ y^{2}[ 1+{{( \frac{dy}{dx} )}^{2}} ]=k^{2} $
Therefore $ y^{2}+y^{2}{{( \frac{dy}{dx} )}^{2}}=k^{2} $
Therefore $ y^{2}{{( \frac{dy}{dx} )}^{2}}=k^{2}-y^{2} $ .
BETA
