If y = y ( x ) is the solution curve of the differential equation x...

${x}^{2}dy + ydx = \dfrac{dx}{x}$

$⇒ \dfrac{dy}{dx} + \dfrac{y}{{x}^{2}} = \dfrac{1}{{x}^{3}}$

$I . F = {e}^{∫ \dfrac{1}{{x}^{2}}dx} = {e}^{- \dfrac{1}{x}}$

$⇒ y \cdot {e}^{- \dfrac{1}{x}} = ∫ {e}^{- \dfrac{1}{x}} \cdot \dfrac{1}{{x}^{3}}dx + C$

$\text{Let}\dfrac{- 1}{x} = t ⇒ \dfrac{1}{{x}^{2}}dx = dt$

$⇒ y \cdot {e}^{- \dfrac{1}{x}} = ∫ - t{e}^{t} \cdot dt + C$

$= - \left[t{e}^{t} - {e}^{t}\right] + C$

$⇒ y \cdot {e}^{- \dfrac{1}{x}} = \dfrac{1}{x}{e}^{- \dfrac{1}{x}} + {e}^{- \dfrac{1}{x}} + C$

$\text{Put}x = 1$

$⇒ ( 1 ) \cdot {e}^{- 1} = \dfrac{{e}^{- 1}}{1} + {e}^{- 1} + C$

$⇒ C = - {e}^{- 1}$

Equation is $y \cdot {e}^{- \dfrac{1}{x}} = \dfrac{1}{x}{e}^{- \dfrac{1}{x}} + {e}^{- \dfrac{1}{x}} - {e}^{- 1}$

$⇒ y = \dfrac{1}{x} + 1 - \dfrac{e\dfrac{1}{x}}{e}$

$\text{At}x = \dfrac{1}{2} ⇒ y\left(\dfrac{1}{2}\right) = 2 + 1 - \dfrac{{e}^{2}}{e} ⇒ y = 3 - e$