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If $y=x+\frac{1}{x}$, then
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Verified Answer
The correct answer is:
$x^2 \frac{d y}{d x}-x y+2=0$
$y=x+\frac{1}{x} \Rightarrow \frac{d y}{d x}=1-\frac{1}{x^2}$
Therefore,
$x^2 \cdot \frac{d y}{d x}-x y+2=x^2\left(1-\frac{1}{x^2}\right)-x\left(x+\frac{1}{x}\right)+2=0$
Therefore,
$x^2 \cdot \frac{d y}{d x}-x y+2=x^2\left(1-\frac{1}{x^2}\right)-x\left(x+\frac{1}{x}\right)+2=0$
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