Differential Equations Question 136
Question: The differential equation satisfied by the family of curves $ y=ax\cos ( \frac{1}{x}+b ) $ , where a, b are parameters, is
[MP PET 2003]
Options:
A) $ x^{2}y_2+y=0 $
B) $ x^{4}y_2+y=0 $
C) $ xy_2-y=0 $
D) $ x^{4}y_2-y=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ y=ax\cos ( \frac{1}{x}+b ) $
…….. (i) Differentiate (i), we get $ y_1=a[ \cos ( \frac{1}{x}+b )-x\sin ( \frac{1}{x}+b )( \frac{-1}{x^{2}} ) ] $
$ =a[ \cos ( \frac{1}{x}+b )+\frac{1}{x}\sin ( \frac{1}{x}+b ) ] $ …..(ii) Again, differentiate (ii), we get $ y_2=\frac{-a}{x^{3}}\cos ( \frac{1}{x}+b ) $
$ =\frac{-ax}{x^{4}}\cos ( \frac{1}{x}+b ) $
$ =\frac{-y}{x^{4}} $
Therefore $ ( \frac{1}{y^{2}}-\frac{1}{y} )dy+( \frac{1}{x^{2}}+\frac{1}{x} )dx=0 $ .
BETA
