Question

If $y=x+\frac{1}{x}$, then

• A. $x^2\left(\frac{dy}{dx}\right)+xy=0$
• B. $x^2\left(\frac{dy}{dx}\right)+xy+2=0$
• C. $x^2\left(\frac{dy}{dx}\right)-xy+2=0$
• D. None of these.

Answer 1

The correct option is C

$x^2\left(\frac{dy}{dx}\right)-xy+2=0$

Explanation for the correct option.

$$\begin{gathered} y=x+\frac{1}{x} \\ \Rightarrow \frac{dy}{dx}=1-\frac{1}{x^2} \\ \Rightarrow \frac{dy}{dx}=\frac{x^2-1}{x^2} \\ \Rightarrow x^2\left(\frac{dy}{dx}\right)=x^2-1 \end{gathered}$$

Now, by substituting the value of $x$ and $y$ in $x^2\left(\frac{dy}{dx}\right)-xy+2=0$, we get

$\begin{array}{rcl}{x}^2-1-x\left(x+\frac{1}{x}\right)+2& =& 0\\ \Rightarrow x^2-1-x^2-1+2& =& 0\\ \Rightarrow 0& =& 0\end{array}$

Hence, option C is correct.

Explanation for the incorrect options.

Option A

By substituting the value of $x$ and $y$ in $x^2\left(\frac{dy}{dx}\right)+xy=0$, we get

$\begin{array}{rcl}{x}^2-1-x\left(x+\frac{1}{x}\right)& =& 0\\ \Rightarrow x^2-1-x^2-1& =& 0\\ \Rightarrow -2& \ne & 0\end{array}$

Hence, option A is incorrect.

Option B

By substituting the value of $x$ and $y$ in $x^2\left(\frac{dy}{dx}\right)+xy+2=0$, we get

$\begin{array}{rcl}{x}^2-1+x\left(x+\frac{1}{x}\right)+2& =& 0\\ \Rightarrow x^2-1+x^2+1& =& 0\\ \Rightarrow 2x^2& \ne & 0\end{array}$

Hence, option B is incorrect.

Hence, option C is correct.