Question

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

• A. $x^2{y}_2+y=0$
• B. $x^4{y}_2+y=0$
• C. $x{y}_2-y=0$
• D. $x^4{y}_2-y=0$

Answer 1

Answer: (B)

Given that, $y=ax\cos \left(\frac{1}{x}+b\right)$
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
$y_2=a[-\sin \left(\frac{1}{x}+b\right)\left(-\frac{1}{x^2}\right)-\frac{1}{x^2}\sin \left(\frac{1}{x}+b\right)$
$-\frac{1}{x^3}\cos \left(\frac{1}{x}+b\right)]$
$\Rightarrow y_2=-\frac{ax}{x^4}\cos \left(\frac{1}{x}+b\right)$
$\Rightarrow x^4{y}_2+y=0$