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(A) Given the differential equation: $x^3 \frac{dy}{dx} + xy - 1 = 0$.Rearranging the terms,we get: $\frac{dy}{dx} + \frac{y}{x^2} = \frac{1}{x^3}$.Th

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(A) Given the differential equation: $x^3 \frac{dy}{dx} + xy - 1 = 0$.Rearranging the terms,we get: $\frac{dy}{dx} + \frac{y}{x^2} = \frac{1}{x^3}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x^2}$ and $Q = \frac{1}{x^3}$.The integrating factor $(IF)$ is $e^{\int P dx} = e^{\int \frac{1}{x^2} dx} = e^{-\frac{1}{x}}$.The general solution is $y \cdot IF = \int Q \cdot IF dx + C$.$y \cdot e^{-\frac{1}{x}} = \int \frac{1}{x^3} e^{-\frac{1}{x}} dx$.Let $t = -\frac{1}{x}$,then $dt = \frac{1}{x^2} dx$. Also,$\frac{1}{x} = -t$.Substituting these into the integral: $y \cdot e^{-\frac{1}{x}} = \int (-t) e^t dt = -(t e^t - e^t) + C = e^t(1 - t) + C$.$y \cdot e^{-\frac{1}{x}} = e^{-\frac{1}{x}}(1 + \frac{1}{x}) + C$.Dividing by $e^{-\frac{1}{x}}$,we get $y = 1 + \frac{1}{x} + C e^{\frac{1}{x}}$.Given $y(\frac{1}{2}) = 3 - e$,we have $3 - e = 1 + \frac{1}{1/2} + C e^{\frac{1}{1/2}} = 1 + 2 + C e^2 = 3 + C e^2$.Thus,$3 - e = 3 + C e^2 \implies C e^2 = -e \implies C = -\frac{1}{e} = -e^{-1}$.So,$y(x) = 1 + \frac{1}{x} - e^{-1} e^{\frac{1}{x}} = 1 + \frac{1}{x} - e^{\frac{1}{x} - 1}$.For $x = 1$,$y(1) = 1 + \frac{1}{1} - e^{\frac{1}{1} - 1} = 1 + 1 - e^0 = 2 - 1 = 1$.

Let $y=y(x)$ be the solution of the differential equation $x^3 dy + (xy - 1) dx = 0, x > 0$,with $y(\frac{1}{2}) = 3 - e$. Then $y(1)$ is equal to

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