If \(y=y(x)\) is the solution of the differential equation \(x \frac{\mathrm{dy}}{\mathrm{d} x}+2 y=x^2\) satisfying \(y(1)=1\), then the value of \(y\left(\frac{1}{2}\right)\) is

\(\begin{aligned} & x \frac{\mathrm{~d} y}{\mathrm{~d} x}+2 y=x^2 \\ & \frac{\mathrm{~d} y}{\mathrm{~d} x}+\left(\frac{2}{x}\right) y=x \\ & \text { I.F. }=\mathrm{e}^{\int \frac{2}{x} \mathrm{~d} x}=\mathrm{e}^{2 \log x}=x^2\end{aligned}\)
\(\therefore \quad\) Solution of differential equation is
\(\begin{aligned}
& \quad y \cdot x^2=\int x \cdot x^2 \mathrm{~d} x+\mathrm{c} \\
& \quad y x^2=\frac{x^4}{4}+\mathrm{c} ...(i)\\
& \therefore \quad \\
& y=\frac{x^2}{4}+\frac{\mathrm{c}}{x^2}
...(ii)\end{aligned}\)
\(\begin{aligned} & \text { Given, } y(1)=1 \\ & \Rightarrow 1=\frac{1}{4}+\mathrm{c} \\ & \Rightarrow \mathrm{c}=\frac{3}{4}\end{aligned}\)
\(\therefore \quad\) equation (i) becomes,
\(\begin{aligned}
& y=\frac{x^2}{4}+\frac{3}{4 x^2} \\
\therefore \quad & y\left(\frac{1}{2}\right)=\frac{\left(\frac{1}{2}\right)^2}{4}+\frac{3}{4 \times\left(\frac{1}{2}\right)^2}=\frac{1}{16}+3=\frac{49}{16}
\end{aligned}\)