If $y=y(x)$ is the solution curve of the differential equation $x^2 d y+\left(y-\frac{1}{x}\right) d x=0 ; x0$ and $y(1)=1$, then $y\left(\frac{1}{2}\right)$ is equal to: [JEE Main 2021, 1 Sep (Shift 2)]

Question

If $y=y(x)$ is the solution curve of the differential equation $x^2 d y+\left(y-\frac{1}{x}\right) d x=0 ; x>0$ and $y(1)=1$, then $y\left(\frac{1}{2}\right)$ is equal to: [JEE Main 2021, 1 Sep (Shift 2)]

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If \(y=y(x)\) is the solution curve of the differential equation \(x^2 d y+\left(y-\frac{1}{x}\right) d x=0 ; x>0\) a…

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If \(y=y(x)\) is the solution curve of the differential equation \(x^2 d y+\left(y-\frac{1}{x}\right) d x=0 ; x&gt;0\) a…

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If \(y=y(x)\) is the solution curve of the differential equation \(x^2 d y+\left(y-\frac{1}{x}\right) d x=0 ; x>0\) a…