Explanation

(b)
Given, $y = ax\cos\left( \frac{1}{x} + b \right)$
$\Rightarrow$ $y_{1} = - ax\sin{\left( \frac{1}{x} + b\ \right) \times}\left( - \frac{1}{x^{2}} \right) + a\cos{(\frac{1}{x}} + b)$
$\Rightarrow$ $y_{1} = \frac{a}{x}\sin\left( \frac{1}{x} + b \right) + {acos}\left( \frac{1}{x} + b \right)$
$\Rightarrow$ $xy_{1} = {asin}{\left( \frac{1}{x} + b \right) + y}$
$\Rightarrow$ $y_{1} + xy_{2} = {acos}{\left( \frac{1}{x} + b \right)\left( - \frac{1}{x^{2}} \right) + y_{1}}$
$\Rightarrow$ $x^{3}y_{2} = - {acos}\left( \frac{1}{x} + b \right)$
$\Rightarrow$ $x^{4}y_{2} + y = 0$