Question

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

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

Answer 1

Answer: (B)

Given that, $y=a x \cos \left(\frac{1}{x}+b\right)$
On differentiating w.r.t. $x$, we get
$y_{1}=a\left[\cos \left(\frac{1}{x}+b\right)-x \sin \left(\frac{1}{x}+b\right)\left(-\frac{1}{x^{2}}\right)\right]$
$\Rightarrow y_{1}=a\left[\cos \left(\frac{1}{x}+b\right)+\frac{1}{x} \sin \left(\frac{1}{x}+b\right)\right]$
Again differentiating, we get
$y_{2}=a\left[-\sin \left(\frac{1}{x}+b\right)\left(-\frac{1}{x^{2}}\right)-\frac{1}{x^{2}} \sin \left(\frac{1}{x}+b\right)\right.$
$\left.-\frac{1}{x^{3}} \cos \left(\frac{1}{x}+b\right)\right]$
$\Rightarrow y_{2}=-\frac{a x}{x^{4}} \cos \left(\frac{1}{x}+b\right)$
$\Rightarrow x^{4} y_{2}+y=0$