If y = y ( x ) is the solution curve of the differential equation x 2   d y + y - 1 x d x = 0 ;  …

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

$\Rightarrow \frac{dy}{dx}+\frac{y}{x^2}=\frac{1}{x^3}$

$\mathrm{I}.\mathrm{F}=e^{\int \frac{1}{x^2}dx}=e^{-\frac{1}{x}}$

$\Rightarrow y·e^{-\frac{1}{x}}=\int e^{-\frac{1}{x}}·\frac{1}{x^3}dx+C$

$\text{ Let }\frac{-1}{x}=t\Rightarrow \frac{1}{x^2}dx=dt$

$\Rightarrow y·e^{-\frac{1}{x}}=\int -t{e}^t·dt+C$

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

$\Rightarrow y·e^{-\frac{1}{x}}=\frac{1}{x}{e}^{-\frac{1}{x}}+e^{-\frac{1}{x}}+C$

$\text{ Put }x=1$

$\Rightarrow \left(1\right)·e^{-1}=\frac{e^{-1}}{1}+e^{-1}+C$

$\Rightarrow \mathrm{C}=-{\mathrm{e}}^{-1}$

Equation is $y·e^{-\frac{1}{x}}=\frac{1}{x}{e}^{-\frac{1}{x}}+e^{-\frac{1}{x}}-e^{-1}$

$\Rightarrow y=\frac{1}{x}+1-\frac{e^{\frac{1}{x}}}{e}$

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