Q68 / 200 · Manipal (MET) 2011 · Mathematics

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
    $x^2 y_2+y=0$
  • Correct answer
    B
    $x^4 y_2+y=0$
  • C
    $x y_2-y=0$
  • D
    $x^4 y_2-y=0$

Solution

Given that, $y=a x \cos \left(\frac{1}{x}+b\right)$ On differentiating w.r.t. $x$, we get $\begin{aligned} y_1 & =a\left[\cos \left(\frac{1}{x}+b\right)-x \sin \left(\frac{1}{x}+b\right)\left(-\frac{1}{x^2}\right)\right] \\ \Rightarrow y_1 & =a\left[\cos \left(\frac{1}{x}+b\right)+\frac{1}{x} \sin \left(\frac{1}{x}+b\right)\right]\end{aligned}$ Again differentiating, we get $\begin{aligned} & y_2=a\left[-\sin \left(\frac{1}{x}+b\right)\left(-\frac{1}{x^2}\right)-\frac{1}{x^2} \sin \left(\frac{1}{x}+b\right)\right. \\ & \left.-\frac{1}{x^3} \cos \left(\frac{1}{x}+b\right)\right] \\ & \Rightarrow \quad y_2=-\frac{a x}{x^4} \cos \left(\frac{1}{x}+b\right) \\ & \Rightarrow \quad x^4 y_2+y=0\end{aligned}$

Hence, the correct answer is option: 'B'.

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