If y=y(x) is the solution curve of the differ | vedclass

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(D) The given differential equation is $x^{2} dy + (y - \frac{1}{x}) dx = 0$.Dividing by $x^{2} dx$,we get $\frac{dy}{dx} + \frac{y}{x^{2}} = \frac{1}

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$3 - e$

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(D) The given differential equation is $x^{2} dy + (y - \frac{1}{x}) dx = 0$.Dividing by $x^{2} dx$,we get $\frac{dy}{dx} + \frac{y}{x^{2}} = \frac{1}{x^{3}}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x^{2}}$ and $Q(x) = \frac{1}{x^{3}}$.The integrating factor $IF = e^{\int P(x) dx} = e^{\int \frac{1}{x^{2}} dx} = e^{-\frac{1}{x}}$.The solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$.$y e^{-\frac{1}{x}} = \int \frac{1}{x^{3}} e^{-\frac{1}{x}} dx + C$.Let $t = -\frac{1}{x}$,then $dt = \frac{1}{x^{2}} dx$. Also $\frac{1}{x} = -t$.$y e^{-\frac{1}{x}} = \int (-t) e^{t} dt + C = -(t e^{t} - e^{t}) + C = e^{t}(1 - t) + C$.Substituting $t = -\frac{1}{x}$,we get $y e^{-\frac{1}{x}} = e^{-\frac{1}{x}}(1 + \frac{1}{x}) + C$.Given $y(1) = 1$,so $1 \cdot e^{-1} = e^{-1}(1 + 1) + C \implies e^{-1} = 2e^{-1} + C \implies C = -e^{-1}$.Thus,$y e^{-\frac{1}{x}} = e^{-\frac{1}{x}}(1 + \frac{1}{x}) - e^{-1}$.For $x = \frac{1}{2}$,$y e^{-2} = e^{-2}(1 + 2) - e^{-1} = 3e^{-2} - e^{-1}$.Dividing by $e^{-2}$,we get $y = 3 - e^{-1} \cdot e^{2} = 3 - e$.

If $y=y(x)$ is the solution curve of the differential equation $x^{2} dy + (y - \frac{1}{x}) dx = 0$ for $x > 0$ and $y(1) = 1$,then $y(\frac{1}{2})$ is equal to:

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