Question

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

• A. $x ^ { 2 } y _ { 2 } + y = 0$
• B. $x ^ { 4 } y _ { 2 } + y = 0$
• C. $x y _ { 2 } - y = 0$
• D. $x ^ { 4 } y _ { 2 } - y = 0$

Answer 1

Answer: (B)

$x ^ { 4 } y _ { 2 } + y = 0$

Answer 2

$y = a x \cos \left( \frac { 1 } { x } + b \right)$ .…. (i)

Differentiate (i), we get

$y _ { 1 } = a \left[ \cos \left( \frac { 1 } { x } + b \right) - x \sin \left( \frac { 1 } { x } + b \right) \left( \frac { - 1 } { x ^ { 2 } } \right) \right]$

$= a \left[ \cos \left( \frac { 1 } { x } + b \right) + \frac { 1 } { x } \sin \left( \frac { 1 } { x } + b \right) \right]$ .....(ii)

Again, differentiate (ii), we get

$y _ { 2 } = \frac { - a } { x ^ { 3 } } \cos \left( \frac { 1 } { x } + b \right)$

$= \frac { - a x } { x ^ { 4 } } \cos \left( \frac { 1 } { x } + b \right)$ $= \frac { - y } { x ^ { 4 } }$ ⇒ $x ^ { 4 } y _ { 2 } + y = 0$ .