$\frac{d y}{d x}+\frac{y}{x}=x^2$
On comparing to $\frac{d y}{d x}+P(x) y=Q(x)$
$$
\begin{aligned}
& P(x)=\frac{1}{x}, Q(x)=x^2 \\
& \mathrm{IF}=e^{\int P d x}=e^{\int \frac{1}{x} d x}=e^{\ln x}=x \\
& y \cdot(\mathrm{IF})=\int Q(x)(\mathrm{IF}) d x+C \\
& y \cdot x=\int x^2 x d x+C=\frac{x^4}{4}+C \\
& y=\frac{x^3}{4}+\frac{C}{x} \\
& y(2)=\frac{2^3}{4}+\frac{C}{2}=2+\frac{C}{2} \\
& y(1)=\frac{1}{4}+\frac{C}{1}=\frac{1}{4}+C \\
& 2 y(2)-y(1)=2\left(2+\frac{C}{2}\right)-\left(\frac{1}{4}+C\right) \\
&=4+C-\frac{1}{4}-C=\frac{15}{4}
\end{aligned}
$$