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(B) Given the family of curves: $y = ax \cos \left( \frac{1}{x} + b \right) \dots (i)$Differentiating with respect to $x$ using the product rule:$y_1

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$x^4 y_2 + y = 0$

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(B) Given the family of curves: $y = ax \cos \left( \frac{1}{x} + b \right) \dots (i)$Differentiating with respect to $x$ using the product rule:$y_1 = a \cos \left( \frac{1}{x} + b \right) + ax \left( -\sin \left( \frac{1}{x} + b \right) \right) \left( -\frac{1}{x^2} \right)$$y_1 = a \cos \left( \frac{1}{x} + b \right) + \frac{a}{x} \sin \left( \frac{1}{x} + b \right) \dots (ii)$Differentiating again with respect to $x$:$y_2 = a \left( -\sin \left( \frac{1}{x} + b \right) \right) \left( -\frac{1}{x^2} \right) + a \left( -\frac{1}{x^2} \right) \sin \left( \frac{1}{x} + b \right) + \frac{a}{x} \cos \left( \frac{1}{x} + b \right) \left( -\frac{1}{x^2} \right)$$y_2 = \frac{a}{x^2} \sin \left( \frac{1}{x} + b \right) - \frac{a}{x^2} \sin \left( \frac{1}{x} + b \right) - \frac{a}{x^3} \cos \left( \frac{1}{x} + b \right)$$y_2 = -\frac{a}{x^3} \cos \left( \frac{1}{x} + b \right)$Multiply both sides by $x^4$:$x^4 y_2 = -ax \cos \left( \frac{1}{x} + b \right)$Since $y = ax \cos \left( \frac{1}{x} + b \right)$,we have:$x^4 y_2 = -y$$x^4 y_2 + y = 0$

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

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